Preparing for SMO Junior Exam Success

hello everyone you may be pretty nervous right now as the Smo junior is approaching and so you're looking for some last minute ideas that can help you to score one or two more points hopefully some of the things we're going to discuss here will be useful but take note that the whole point of olympiads is that the questions are not predictable so there's really no way that you can guarantee a type of question will appear or will not appear but nonetheless I think there's a use to discussing some frequently occurring Concepts not just for scoring a couple more marks but just for helping you realize that these problems are actually approachable and even for those of you who are pretty seasoned one or two of these ideas may be a little fresh now one of the things that I found very interesting about olympiads is just all the very big numbers now it felt kind of cool to be able to deal with huge numbers but the trick is that you don't really deal with the huge numbers for example if we look at the first problem you have got huge powers but what you do when you're handling huge Powers is normally that you want to either make the base the same or you want to make the power the same now here you can see that the powers are 300 200 and 100. so all of them are something hundred and I can write this all As something to the power of 100 Z is six to the power of 100 why has 100 pairs of trees so it's 3 squared and you have 100 copies there's 200 copies of three which is 9 to the power of one hundred X is 2 to the power of 300 and so that has 100 triples of tools and straight away we can compare this very easily it means that Y is bigger than X is bigger than Z so what I'm going to do for this video is that you see there's a second question below I'm not going to give you the solution to the second question I'll only just give the idea and if you want to pause the video and work through it you're most welcome too and discuss your solution in the comments below or ask for help if you need to in this case for the second question what you want to notice is that the idea is the same since 4026 is just 2 times of 2013 which is the other power on the right hand side remember that for the Smo every constant is there for a reason yes you may see 2023 all over the place but 2023 appearing more than once it's probably there for a good reason continuing on this theme of big numbers one of the fun things about SML problems is that you don't necessarily have to even understand why it turns out to be a nice number so one option whenever you encounter a question like this is always to find a pattern if 999 squared plus one nine nine nine nine is supposed to have a nice square root well I don't see why four nines is so important it probably works with one or two nine so why not just try to look for the small ones which you can calculate nine squared is 81 plus 19 is 100 the square root of 100 is 10. we probably at least want one more so 99 squared plus 199 if you were to calculate it 9801 plus 199 is ten thousand so this is 100. and I think you can roughly tell what's the pattern right the pattern is that nine squared plus something square rooted you get 10 99 squared plus something you square root it you get 100 so it is not too surprising that based on the pattern the answer should be ten thousand no using a bit of algebra however is actually the intended method and this one has a really obvious pattern but not all the time are you going to get an obvious pattern such as in the second example below so to give an example of how to use the algebra you usually just look at the most frequently occurring number or the most important looking number which in this case is 9999 and called it X usually the reason why this even turn out to be a nice number is that there is some reason that comes from a bit of algebra now 199999 you would like to also write this in a way that is related to X so don't write it as 10 000 plus X you want to make it as related to X as possible and 10 000 is just one more than 999 so instead write it as 2x Plus 1. you see that this is very familiar looking any of you that have even done a little bit of algebra are likely to recognize this as X Plus 1 squared and so that is why you get 10 000. so many times algebra is tested using so-called numerical questions and you have to recognize that there is this particular algebraic Identity or equation that you can use so for instance for the example you would like to use a difference of squares factorization remember that M squared minus N squared is M plus n times M minus n this is your difference of squares factorization and so I would suggest combining the first and third and the second and fourth if you're wondering why that combination Well it can't be the first two because a difference of square s requires something minus something so let's take the first center and the second and fourth I'll leave you to work out this question by the way all of these questions are passed Smo problems so these are real these are the kind of questions you will see sometimes but even if you don't see a question that looks very much like this the idea is still important we talk about idea questions that involve integers is one of those where there isn't really a standard method but you have to realize that a question that says positive integers is actually information because normally all that we have are just real numbers not positive integers how does that change things compared to our usual equations well you notice here that we have got for the first question two equations for three unknowns and for the second question we have only got one equation for three unknotes so it would seem that you don't have enough information but with positive integers it just means that there are limited possibilities just based on you being told that there are positive integers for example if two positive integers add up to five then it can either be one and four or two and three if I told you two real numbers add up to five oh that's quite a lot of possibilities isn't it so what we do here now usually you are looking for some reason why the numbers are restricted and cannot be too big the second equation doesn't tell me that much I'm going to focus on the first equation and I will just multiply the two throughout on the right hand side and I'm going to shift the 2xy over I'm going to shift it into a place that would help you to recognize what's going on this is a good friend x minus y squared now you might be thinking well what's the point of this why are we doing all this work well a square number plus a square number is two now I say square number because x squared doesn't have to be a square number right if x is 1 3 then 1 9 it's not something you would call a square number but here because they're integers they really are square numbers in the traditional sense so obviously this can only happen if both of them are 1. the implication being that Z is one x minus y is 1 or negative 1. of course at this point we just have a usual pair of simultaneous equations with Z being 1 X Plus Y is 2021 and x minus y is one or x minus y equals to negative 1 and it's not very difficult to see that X Y would be 10 10 11 or vice versa so the question asks you for the value of X1 plus X2 which are the two possible values of X that would just be 10 10 plus 10 11 which is 20 21. using the same idea you can look at the second question and just think you know what fractions are usually quite small if it's 1 over some whole number if let's say that all of them are even just five one-fifth plus one-fifth plus one-fifth doesn't even reach one it's not even close to one it's three-fifths so you need to have a b and c to be quite small otherwise the fractions are too small and where you would start is to say that well one over a has to be at least one third because if it's one quarter one quarter one quarter then you don't even reach one so you will start off by saying that this is at least one third and so a can only be two or three and from there I think things are relatively easy one more thing before we move on away from talking about equations not everything that has an equal sign is an equation now you may have seen some of these kinds of questions in the recent smos and think oh my goodness how does anyone solve that equation then you are kind of asking the wrong question I've written in the header identities and equations now equation means equal if blah blah blah blah blah now identities that means they are identical so always the same it's just like in English right if I say that drawn is Peter's brother now um John is Peter's brother doesn't change right Johnny is always Peter's brother if Peter flies away to the U.S John is still Peter's brother if John gets into it John is still Peter's brother so that is an identity it's always the same it always is true whereas if I told you that John is eating lunch now that is is talking about right now John is eating lunch Jon isn't eating lunch for his entire life continuously and doing nothing else John is eating lunch only at the moment whenever you're solving equations it's usually something that's at the moment so in our previous example X Plus y plus Z is 2022 in this problem only one over a plus one over B plus one over C is one in this problem only that's not true for our two examples here when we say equals what we're trying to say in this case is that we have this now it's not asking you to solve we are saying assume that now if I tell you that these are the same it means that they are literally the same in the same way as if I told you that for Pythagoras Theorem c squared equals to a squared plus b squared this is not asking you to solve it it's asking you to use it so you can use this on any triangle any right angle triangle I should say so here you can put in any value of x that you want and this will be true now this is not cheating this is not one of those cases where oh the answer is unique so I'm assuming it no we are trying to say that those two things are literally the same thing if I told you that X is 2x divided by 2. those are identical I can put in any X I like so all you need to do is to choose the suitable number X in order for the expression to appear a0 minus A1 Plus A2 minus A3 plus A4 minus A5 that's all somewhere down here and you would like to make sure that when you put in a number it looks like that so for this case we would let X to be negative one if x is negative one the left hand side is just one in all of these brackets which we are very happy with whereas the right hand side is going to be exactly what we want just in the opposite order so it would be negative 1 to odd Powers is negative negative 1 to the power of even Powers is going to be positive one so you see that this is what we want just in the reverse order and so your answer is simply 18. can't do the same idea for the second example the value of x will be a bit different and also just pay attention to the fact that a0 is missing so as you're solving that make sure that you don't forget attention to detail is really important for Olympians because these are questions you have never seen before so it's very easy to overlook something so read it very carefully and also read what they ask you to find very carefully unfortunately there are no partial marks so if you are 90 correct you do not get 90 of the mark you get no marks and that's really unfortunate at a Midway point we have 10 things we want to say so in the Midway point I show you another very scary looking thing that is not really scary in a very specific circumstance now when you see this square root is inside square roots I'm sure it looks horrible but what is different about this from normal square root questions is that very often you're going to be given the same thing except with a sign flipped from plus to minus and minus to plus we call these conjugates for example if I have got a 9 plus square root 3 and 9 minus square root 3. so you flip the sign on the square root those are called conjugates now why do we like conjugates well conjugates sort of is another word for partners and partners are supposed to work well together so in what way do they work well well in that example here the sum would be nice because the square root is gone the product is also nice because if you use your difference of squares expansion the sometimes the difference is going to put a square on the square root which is what we always want to do with square roots isn't it so just an illustration there of why this conjugates work well together you might be saying that so what I mean I can't add or subtract or multiply or divide this stuff they're all in just separate giant square roots well not to worry what you do is that first of all just call this whole thing let's say x we've done that before let's do it again and I'm going to square it very carefully just a quick note that for this question you will see this brackets that have this sort of L shapes and reverse L shapes this is called a flaw and flaw just means round down remember the Smo questions always have a whole number answer and so sometimes these are put there just to mean round down so don't worry about it this is just for your final answer okay so x squared is going to be 45 plus square root 2021 from squaring the first thing 45 minus square root 2021 from squaring the second thing and then remember that a minus B squared this is a squared minus 2 AV plus b squared so please don't forget the middle term somehow we always tend to forget don't forget it so it will be -2 times the first one times the second one which I can just merge into a single square root now remember what we said about conjugates they go well together by squaring it we are kind of trying to remove this Berry kit and allow them to meet so by them meeting together the sum you get rid of the square root which is 90. for the other one the product 45 plus root 2021 45 minus root 2021 just gives you 45 squared minus 2021 which you will be happy to know 45 squared minus 2021 is just four and so this is going to be equal to 86. that's not our answer right because remember this is X squared so x squared is 86. and therefore what's the flow of X the flow of X means that X is nine point something since 9 square is 81 and 10 square is 100 you round it down to 9. well if you haven't seen my video uh just recently on the Smo tips do take a look because a question like this is also an example where you can exploit a little bit of estimation because square root of 2021 is also very close to 45 and so another way to do this is to Simply take the second one as almost nothing and X is going to be just roughly square root of 90 which is also nine point something so if you recall what I mentioned in the other video use the fact that you don't have to show your workings and use the fact that the answers are whole numbers sometimes you can just do some estimations if you are stuck and are unable to find the proper method you can do something similar for the second example but I would encourage you to try to do it using the intended method I've shown just in case some square roots are a little bit more difficult to estimate than the one that I've shown here on the topic of square roots one of the places where you will see square roots appearing is Pythagoras Theorem Pythagoras Theorem is very useful for the geometry questions in the Smo Junior mainly because that's the point of difference from primary school geometry Olympics to secondary school geometry in olympiads you might be thinking well what's the big deal I've used Pythagoras theorem in primary school before what's the difference the answer is very simple in Primary School you have that assurance that everything is a whole number for the Smo you only have an assurance that The Final Answer is a whole number and so you want to use Pythagoras but use it carefully try not to make a glut of square roots appear too early by always writing c squared a squared plus b squared you don't want your square roots because that's not nice we don't voluntarily bombard ourselves with square roots so for the first question you are told that this is a piece of paper and you have folded it over before I use the Pythagoras Theorem I'm just going to note the lengths that we're given this is 12 it's a rectangle so this is also 12. this is 24 and he got folded over to here so that's also 24. now I mean to use Pythagoras not in the sense of okay um AC is square root 12 squared plus 24 squared no this is not what I mean by using Pythagoras it's not wrong but it's not useful is it it has nothing to do with the area of the overlapping region so when you're using Pythagoras you want to be a little bit more judicious and ask yourself well what am I trying to find to find the area of the Shaded region what you would like to find is this length here because that would be the base of the Shaded triangle with the height being CD which is 12. therefore if you would like to find that length what have we done so far in all of these issues when we're trying to find something and we can't find it we call it X we call it X because that's what we want to find and now I can also call this 24 minus X and one more thing is that you want to use as many nice properties as you can so that you don't end up with a huge number of equations hopefully you realize that just by symmetry the two unshaded triangles namely the tool that I'm shading right now are congruent to each other meaning that they are the same size same shape and so therefore I could also shift the 24 minus X here or I could shift the X there all the same right it's just by symmetry because of that symmetry you realize that we only have one unknown and one equation which is going to be over here 12 squared plus 24 minus x squared equals x squared and this is how you use Pythagoras you want to use Pythagoras without just waiting for numerical values because Pythagoras is just a free equation you may not like equations every very much but that's how math works you want more equations not less equations so let's try to just solve this I will expand 24 squared minus 2 times 24 times x plus x squared and just nice x squared cancels out so after solving this X is equal to 15. and that is the length you want the area is half times 12 times 15. so you've seen how Pythagoras can help you to find lengths even if you seem to say I only know one length how can I use one line to find two lengths you can use one length to find two related legs and so I would ask you to try the same thing for the second question where the idea is to make this right angle triangles appear yourself and use symmetry to help you a circle is always symmetric so you get some Symmetry and I'm not just talking about aob and cob being symmetric I can even just add in a line of symmetry for this triangle and that would give me a new triangle with length 7 and a right angle so I can call this height H and the radius r do the same thing here so that you get a rectangle this is seven this is Hitch and this little bit here is the radius minus seven and so you will have two equations seven hit and R as well as the tiny one over there hitch R minus seven and six from here gas and track is not entirely wrong but the easiest way is to just subtract the two equations and then solve for R right that's what you're supposed to find now you notice that we have been making algebra just pop up everywhere algebra and geometry questions algebra in our so-called numerical questions and you may be a little bit sick of algebra here's something that is pretty legitimately not algebra divisibility tests and remainder something that you have seen before in Primary School but I can assure you that there are still harder questions that can be asked than the primary school Olympians an example would be the first one we see over here to use the divisibility test there are two issues one there is no divisibility test for six but that's quite easily resolved six is two times three so just do a test for two and three now two that straightforward the number is odd but for three how do I know the remainder when it is divided by 3 the test for three sum up the digits and then divide by three to get the remainder oh that sounds awful how am I supposed to do that well we can upgrade our divisibility test a little bit and so the upgrade that we can do is that the divisibility test does not just work exactly as it is told to you for example for the test for three or nine you can upgrade it to the sum of just contiguous segments which sounds like an awfully fancy word but what I mean is that you can also split it up anywhere any way you like and add up those things instead of adding it up digit by digit so in order to get the remainder when dividing by three I can do a simpler task which is to add up from 1 to 2021 and when I add up from 1 to 2021 it's not really with the intention of adding up from 1 to 2021 I just want the remainder when that's divided by three so I know how to add up numbers from 1 to 2021 it would be 2021 numbers times the average which is 20 22 over 2. before I calculate this 1011 is already a multiple of three so therefore the remainder when divided by 3 is just 0. but that's not the answer is it because we are not saying that this number is divisible by 6. it is divisible by 3 and it is odd so what can the remainder be when it is divisible by 3 but odd it can't be one because if the remainder is 1 that means that it was not divisible by 3. the only possibility is that it is 3. so know your divisibility tests well and for the second question the same procedure split it into 3 and 11 and for 11. your test is going to be an alternating sum but make sure you know where to alternate from if you are trying to find a remainder let's say if you wanted to find the remainder when 4321 is divided by 11. well if you just did direct division you get 392 remainder 9. so if you said alternating sum and say okay I do 4 minus 3 plus 2 minus 1 is 2 wait that's not right and indeed it is not because for the alternating sum for 11 it needs to be from right to left in other words 1 minus two plus three minus four which gives you negative two and negative 2 means that it is two less than a multiple of 11. and so that's how you get 9 as your remainder so make sure you know your test and these are not just tests for divisibility these are also tests for what the remainder is equal to as I mentioned another comment it's worthwhile to think about why these are true but I do call this a last minute Smo crash course right so I think now is not the time to think about why it is true now is the time to just make sure that you can apply it correctly another word on number Theory this time highest common factor and lowest common multiples if you're like me my first introduction to the highest common factor in primary school is using this table which my teacher called like a hdb block if you want to find let's say the highest common factor of something like 20 and 30 you say okay I take out a tool and then after that I take out a tool again no I can't take out a tour I can take out a five and after that you stop and say that okay there's nothing left already and so my hcf based on the left column is 2 times 5 is 10. and this is probably how your thoughts the first time and this is perfectly good if you know what the numbers are isn't it we don't know what the numbers are for a question like this so I can't even try to draw this hdb block out I want you to have this notion of every number being like a list of Demands every number is like a list of Demands and that List of Demands is based on its prime factorization so here you can guess but I'm just going to show you the full method which is that let's say for 80 and 40 you can prime factorize it as 2 cubed times five and 2 to the power of 4 times 5 12 is 2 squared times 3 and 252 that will take a little while it's 2 squared times 3 squared times seven so each number is a list of Demands and also you can think of it as a list of resources so 80 contains four tools and a five highest common factor is asking you what resources do these two numbers both possess and so this here tells you that X possesses a 2 cubed and a five it does not have a 2 to the power of 4 otherwise that would have been in common for the second statement 2 squared and 2 squared so it means that well X had two cubed so it's okay it also has two squared so if it has three tools it also has two tools however it doesn't have 3 squared it only has one tree otherwise 3 squared would have been in common it also does not have a seven otherwise that would have shown up and so therefore you compare this this it means that the minimum X would need to have two cubed it will need to have a tree and it will need to have a five so that is equal to 120 coincidentally enough it is actually true that 120 is the lowest common multiple of 40 and 12. does it make sense why that's a good idea well the lowest common multiple is kind of the reverse of that so in the second question the lowest common multiple looks for the bigger one it looks for if let's say here 42 is 2 times 3 times 7 and 14 is 2 times 7. why are they asking me for 3 well that means that X must have a 3. and you can reason similarly for the second part so the prime factorization may seem like a little bit of over Q here but I can assure you that you would really like to be able to use the prime factorization for many purposes and one of the purposes that is quite interesting is that you can count factors using the prime factorization for instance if I give you a huge number 2 to the power of 7 times 3 to the power of 11 times 5 to the power of four you would be thinking okay how long is it going to take me to calculate this what if I ask you how many factors does it have well there is actually a very simple shortcut and all you need to do is to take the powers and add one to them and multiply together so seven plus one is eight eleven plus one is twelve four plus one is five and so I can say that this has 480 factors why does that work the reason is actually not so complicated if we continue on this discussion of factors meaning looking for resources for a number to be a factor of this 2 to the seven times three to the power of 11 times 5 to the power of four a factor must look like 2 to the power of something 3 to the power of something and five to the power of something you can't say that my Factor includes the number 19 No 19 is nowhere to be found in that prime factorization we also can say that the factor of 2 cannot exceed 7 because that's as many as is available if you took 2 to the power of 8 that's not a factor and so your options in the power are 0 to 7 0 to 11 and 0 to 4 and yes I mean zero zero means don't use that prime number what's wrong with that one's a factor and one would be 2 to the power of 0 3 to the power of zero and five to the power of zero so nothing wrong with that and 0 to 7 means seven plus one choices 0 to 11 means eleven plus one choices zero to four is four plus one choices and then you multiply them to find the total number of options so you can see here that this is a very useful tool and in fact the questions that I've selected here are relatively tame by comparison because it just says four factors and three devices which is the same as factors but you are asked to find the numbers or to count the numbers that's why you can't just say that I'm just going to list out four factors what's so hard about that because you are asked the reverse question so it helps to know that this is the way to count factors so that you know four factors could either be three plus one or four is two times two which is one plus one times one plus one what does that mean three plus one corresponds to a prime number cubed one plus one and one plus one means that you have two different prime numbers with the power being one so p q where p and Q are prime numbers and with that we just need to count how many numbers are like that the two examples six is two times three is one of the second type and eight equals to two cubed is one of the first time so we just need to list it out that has to be done carefully of course but it is now a systematic task and not just factoring the numbers one by one by one so P cubed the only thing that fits is 3 cubed which is 27 and everything else goes past 50. for PQ is the product of two prime numbers and two times just about everything is going to fit so 2 times 5 7 11 13 17 19 and 23 are all of the prime numbers that can fit that and then you have a handful more three times five three times seven three times eleven three times thirteen three times Seventeen is 51 so we don't include that and then 5 times 7 and that is basically the end of the list everything else is going to exceed and we cannot use a prime squared a prime squared would have three factors not four I think I just bought the second question didn't I so in total here if we count it there are a total of 13 numbers and I've included this question just to mention one thing don't be afraid to do a bit of listing as long as you know what you're doing listing is not some mathematical sin you are allowed to list as long as your list is complete and your list is not going to be 700 numbers long this thing is okay so not everything is going to be some complicated method or some sophisticated theorem and to conclude some things are just common sense you can't really put a finger on a theorem if we look at the first question we might be thinking how am I supposed to actually find that two digit number that is 18 more than the product of its two digits well forget about thinking too hard the answer is 99. because 99 works and obviously you can't find anything larger there are going to be problems like this and problems like these are the thoughts that are challenging those of you who may be a bit too eager to solve the question and then you start cataloging theorem after the theorem of the theorem formula after formula remember these questions are supposed to be fine there are also some others that are just a bit strange looking like the second question if you looked at the second question it doesn't seem very hard to solve because 10 minus 2x is just negative two times of x minus 5. so if I just cancel this off m is going to be negative 2x minus 1. or rather another another way of thinking about it is that X is 1 plus M over negative 2 wait how do I have no Solutions in x so there's some questions which are like this and you just need to look back at your steps and sit back and think what could have gone wrong with my steps how do I find m without finding X but how do I find X when they tell me there's no solution in X so there's no X isn't there no M so this is very confusing and very weird well the only thing that can go wrong is when you cancel off x minus 5. the problem is that what if x equals to five well you might say then don't let it be five precisely so if x equals to 5 is the solution then it is not a solution and so if this was 5 then this is actually going to be not a solution so you want to be able to analyze questions that are unusual that's at the heart of Olympians and so at the end of the day I've just shown you 10 ideas for the assemble Junior but remember just to have some fun and also two reason things out on your own you're all better at math than you think you are so reason things out don't just rely on formulas don't just rely on having seen the question before that's the entire Spirit of doing math olympiads and on that note I'll be concluding this mini crash course feel free to comment below if you have any questions and also feel free to comment below if you'd like to discuss any of the second questions here which I didn't go through in detail so all the best everyone and I hope this helped a little bit

hello everyone you may be pretty nervous right now as the Smo junior is approaching and so you&#39;re looking for some last minute ideas that can help you to score one or two more points hopefully some of the things we&#39;re going to discuss here will be useful but take note that the whole point of olympiads is that the questions are not predictable so there&#39;s really no way that you can guarantee a type of question will appear or will not appear but nonetheless I think there&#39;s a use to discussing some frequently occurring Concepts not just for scoring a couple more marks but just for helping you realize that these problems are actually approachable and even for those of you who are pretty seasoned one or two of these ideas may be a little fresh now one of the things that I found very interesting about olympiads is just all the very big numbers now it felt kind of cool to be able to deal with huge numbers but the trick is that you don&#39;t really deal with the huge numbers for example if we look at the first problem you have got huge powers but what you do when you&#39;re handling huge Powers is normally that you want to either make the base the same or you want to make the power the same now here you can see that the powers are 300 200 and 100. so all of them are something hundred and I can write this all As something to the power of 100 Z is six to the power of 100 why has 100 pairs of trees so it&#39;s 3 squared and you have 100 copies there&#39;s 200 copies of three which is 9 to the power of one hundred X is 2 to the power of 300 and so that has 100 triples of tools and straight away we can compare this very easily it means that Y is bigger than X is bigger than Z so what I&#39;m going to do for this video is that you see there&#39;s a second question below I&#39;m not going to give you the solution to the second question I&#39;ll only just give the idea and if you want to pause the video and work through it you&#39;re most welcome too and discuss your solution in the comments below or ask for help if you need to in this case for the second question what you want to notice is that the idea is the same since 4026 is just 2 times of 2013 which is the other power on the right hand side remember that for the Smo every constant is there for a reason yes you may see 2023 all over the place but 2023 appearing more than once it&#39;s probably there for a good reason continuing on this theme of big numbers one of the fun things about SML problems is that you don&#39;t necessarily have to even understand why it turns out to be a nice number so one option whenever you encounter a question like this is always to find a pattern if 999 squared plus one nine nine nine nine is supposed to have a nice square root well I don&#39;t see why four nines is so important it probably works with one or two nine so why not just try to look for the small ones which you can calculate nine squared is 81 plus 19 is 100 the square root of 100 is 10. we probably at least want one more so 99 squared plus 199 if you were to calculate it 9801 plus 199 is ten thousand so this is 100. and I think you can roughly tell what&#39;s the pattern right the pattern is that nine squared plus something square rooted you get 10 99 squared plus something you square root it you get 100 so it is not too surprising that based on the pattern the answer should be ten thousand no using a bit of algebra however is actually the intended method and this one has a really obvious pattern but not all the time are you going to get an obvious pattern such as in the second example below so to give an example of how to use the algebra you usually just look at the most frequently occurring number or the most important looking number which in this case is 9999 and called it X usually the reason why this even turn out to be a nice number is that there is some reason that comes from a bit of algebra now 199999 you would like to also write this in a way that is related to X so don&#39;t write it as 10 000 plus X you want to make it as related to X as possible and 10 000 is just one more than 999 so instead write it as 2x Plus 1. you see that this is very familiar looking any of you that have even done a little bit of algebra are likely to recognize this as X Plus 1 squared and so that is why you get 10 000. so many times algebra is tested using so-called numerical questions and you have to recognize that there is this particular algebraic Identity or equation that you can use so for instance for the example you would like to use a difference of squares factorization remember that M squared minus N squared is M plus n times M minus n this is your difference of squares factorization and so I would suggest combining the first and third and the second and fourth if you&#39;re wondering why that combination Well it can&#39;t be the first two because a difference of square s requires something minus something so let&#39;s take the first center and the second and fourth I&#39;ll leave you to work out this question by the way all of these questions are passed Smo problems so these are real these are the kind of questions you will see sometimes but even if you don&#39;t see a question that looks very much like this the idea is still important we talk about idea questions that involve integers is one of those where there isn&#39;t really a standard method but you have to realize that a question that says positive integers is actually information because normally all that we have are just real numbers not positive integers how does that change things compared to our usual equations well you notice here that we have got for the first question two equations for three unknowns and for the second question we have only got one equation for three unknotes so it would seem that you don&#39;t have enough information but with positive integers it just means that there are limited possibilities just based on you being told that there are positive integers for example if two positive integers add up to five then it can either be one and four or two and three if I told you two real numbers add up to five oh that&#39;s quite a lot of possibilities isn&#39;t it so what we do here now usually you are looking for some reason why the numbers are restricted and cannot be too big the second equation doesn&#39;t tell me that much I&#39;m going to focus on the first equation and I will just multiply the two throughout on the right hand side and I&#39;m going to shift the 2xy over I&#39;m going to shift it into a place that would help you to recognize what&#39;s going on this is a good friend x minus y squared now you might be thinking well what&#39;s the point of this why are we doing all this work well a square number plus a square number is two now I say square number because x squared doesn&#39;t have to be a square number right if x is 1 3 then 1 9 it&#39;s not something you would call a square number but here because they&#39;re integers they really are square numbers in the traditional sense so obviously this can only happen if both of them are 1. the implication being that Z is one x minus y is 1 or negative 1. of course at this point we just have a usual pair of simultaneous equations with Z being 1 X Plus Y is 2021 and x minus y is one or x minus y equals to negative 1 and it&#39;s not very difficult to see that X Y would be 10 10 11 or vice versa so the question asks you for the value of X1 plus X2 which are the two possible values of X that would just be 10 10 plus 10 11 which is 20 21. using the same idea you can look at the second question and just think you know what fractions are usually quite small if it&#39;s 1 over some whole number if let&#39;s say that all of them are even just five one-fifth plus one-fifth plus one-fifth doesn&#39;t even reach one it&#39;s not even close to one it&#39;s three-fifths so you need to have a b and c to be quite small otherwise the fractions are too small and where you would start is to say that well one over a has to be at least one third because if it&#39;s one quarter one quarter one quarter then you don&#39;t even reach one so you will start off by saying that this is at least one third and so a can only be two or three and from there I think things are relatively easy one more thing before we move on away from talking about equations not everything that has an equal sign is an equation now you may have seen some of these kinds of questions in the recent smos and think oh my goodness how does anyone solve that equation then you are kind of asking the wrong question I&#39;ve written in the header identities and equations now equation means equal if blah blah blah blah blah now identities that means they are identical so always the same it&#39;s just like in English right if I say that drawn is Peter&#39;s brother now um John is Peter&#39;s brother doesn&#39;t change right Johnny is always Peter&#39;s brother if Peter flies away to the U.S John is still Peter&#39;s brother if John gets into it John is still Peter&#39;s brother so that is an identity it&#39;s always the same it always is true whereas if I told you that John is eating lunch now that is is talking about right now John is eating lunch Jon isn&#39;t eating lunch for his entire life continuously and doing nothing else John is eating lunch only at the moment whenever you&#39;re solving equations it&#39;s usually something that&#39;s at the moment so in our previous example X Plus y plus Z is 2022 in this problem only one over a plus one over B plus one over C is one in this problem only that&#39;s not true for our two examples here when we say equals what we&#39;re trying to say in this case is that we have this now it&#39;s not asking you to solve we are saying assume that now if I tell you that these are the same it means that they are literally the same in the same way as if I told you that for Pythagoras Theorem c squared equals to a squared plus b squared this is not asking you to solve it it&#39;s asking you to use it so you can use this on any triangle any right angle triangle I should say so here you can put in any value of x that you want and this will be true now this is not cheating this is not one of those cases where oh the answer is unique so I&#39;m assuming it no we are trying to say that those two things are literally the same thing if I told you that X is 2x divided by 2. those are identical I can put in any X I like so all you need to do is to choose the suitable number X in order for the expression to appear a0 minus A1 Plus A2 minus A3 plus A4 minus A5 that&#39;s all somewhere down here and you would like to make sure that when you put in a number it looks like that so for this case we would let X to be negative one if x is negative one the left hand side is just one in all of these brackets which we are very happy with whereas the right hand side is going to be exactly what we want just in the opposite order so it would be negative 1 to odd Powers is negative negative 1 to the power of even Powers is going to be positive one so you see that this is what we want just in the reverse order and so your answer is simply 18. can&#39;t do the same idea for the second example the value of x will be a bit different and also just pay attention to the fact that a0 is missing so as you&#39;re solving that make sure that you don&#39;t forget attention to detail is really important for Olympians because these are questions you have never seen before so it&#39;s very easy to overlook something so read it very carefully and also read what they ask you to find very carefully unfortunately there are no partial marks so if you are 90 correct you do not get 90 of the mark you get no marks and that&#39;s really unfortunate at a Midway point we have 10 things we want to say so in the Midway point I show you another very scary looking thing that is not really scary in a very specific circumstance now when you see this square root is inside square roots I&#39;m sure it looks horrible but what is different about this from normal square root questions is that very often you&#39;re going to be given the same thing except with a sign flipped from plus to minus and minus to plus we call these conjugates for example if I have got a 9 plus square root 3 and 9 minus square root 3. so you flip the sign on the square root those are called conjugates now why do we like conjugates well conjugates sort of is another word for partners and partners are supposed to work well together so in what way do they work well well in that example here the sum would be nice because the square root is gone the product is also nice because if you use your difference of squares expansion the sometimes the difference is going to put a square on the square root which is what we always want to do with square roots isn&#39;t it so just an illustration there of why this conjugates work well together you might be saying that so what I mean I can&#39;t add or subtract or multiply or divide this stuff they&#39;re all in just separate giant square roots well not to worry what you do is that first of all just call this whole thing let&#39;s say x we&#39;ve done that before let&#39;s do it again and I&#39;m going to square it very carefully just a quick note that for this question you will see this brackets that have this sort of L shapes and reverse L shapes this is called a flaw and flaw just means round down remember the Smo questions always have a whole number answer and so sometimes these are put there just to mean round down so don&#39;t worry about it this is just for your final answer okay so x squared is going to be 45 plus square root 2021 from squaring the first thing 45 minus square root 2021 from squaring the second thing and then remember that a minus B squared this is a squared minus 2 AV plus b squared so please don&#39;t forget the middle term somehow we always tend to forget don&#39;t forget it so it will be -2 times the first one times the second one which I can just merge into a single square root now remember what we said about conjugates they go well together by squaring it we are kind of trying to remove this Berry kit and allow them to meet so by them meeting together the sum you get rid of the square root which is 90. for the other one the product 45 plus root 2021 45 minus root 2021 just gives you 45 squared minus 2021 which you will be happy to know 45 squared minus 2021 is just four and so this is going to be equal to 86. that&#39;s not our answer right because remember this is X squared so x squared is 86. and therefore what&#39;s the flow of X the flow of X means that X is nine point something since 9 square is 81 and 10 square is 100 you round it down to 9. well if you haven&#39;t seen my video uh just recently on the Smo tips do take a look because a question like this is also an example where you can exploit a little bit of estimation because square root of 2021 is also very close to 45 and so another way to do this is to Simply take the second one as almost nothing and X is going to be just roughly square root of 90 which is also nine point something so if you recall what I mentioned in the other video use the fact that you don&#39;t have to show your workings and use the fact that the answers are whole numbers sometimes you can just do some estimations if you are stuck and are unable to find the proper method you can do something similar for the second example but I would encourage you to try to do it using the intended method I&#39;ve shown just in case some square roots are a little bit more difficult to estimate than the one that I&#39;ve shown here on the topic of square roots one of the places where you will see square roots appearing is Pythagoras Theorem Pythagoras Theorem is very useful for the geometry questions in the Smo Junior mainly because that&#39;s the point of difference from primary school geometry Olympics to secondary school geometry in olympiads you might be thinking well what&#39;s the big deal I&#39;ve used Pythagoras theorem in primary school before what&#39;s the difference the answer is very simple in Primary School you have that assurance that everything is a whole number for the Smo you only have an assurance that The Final Answer is a whole number and so you want to use Pythagoras but use it carefully try not to make a glut of square roots appear too early by always writing c squared a squared plus b squared you don&#39;t want your square roots because that&#39;s not nice we don&#39;t voluntarily bombard ourselves with square roots so for the first question you are told that this is a piece of paper and you have folded it over before I use the Pythagoras Theorem I&#39;m just going to note the lengths that we&#39;re given this is 12 it&#39;s a rectangle so this is also 12. this is 24 and he got folded over to here so that&#39;s also 24. now I mean to use Pythagoras not in the sense of okay um AC is square root 12 squared plus 24 squared no this is not what I mean by using Pythagoras it&#39;s not wrong but it&#39;s not useful is it it has nothing to do with the area of the overlapping region so when you&#39;re using Pythagoras you want to be a little bit more judicious and ask yourself well what am I trying to find to find the area of the Shaded region what you would like to find is this length here because that would be the base of the Shaded triangle with the height being CD which is 12. therefore if you would like to find that length what have we done so far in all of these issues when we&#39;re trying to find something and we can&#39;t find it we call it X we call it X because that&#39;s what we want to find and now I can also call this 24 minus X and one more thing is that you want to use as many nice properties as you can so that you don&#39;t end up with a huge number of equations hopefully you realize that just by symmetry the two unshaded triangles namely the tool that I&#39;m shading right now are congruent to each other meaning that they are the same size same shape and so therefore I could also shift the 24 minus X here or I could shift the X there all the same right it&#39;s just by symmetry because of that symmetry you realize that we only have one unknown and one equation which is going to be over here 12 squared plus 24 minus x squared equals x squared and this is how you use Pythagoras you want to use Pythagoras without just waiting for numerical values because Pythagoras is just a free equation you may not like equations every very much but that&#39;s how math works you want more equations not less equations so let&#39;s try to just solve this I will expand 24 squared minus 2 times 24 times x plus x squared and just nice x squared cancels out so after solving this X is equal to 15. and that is the length you want the area is half times 12 times 15. so you&#39;ve seen how Pythagoras can help you to find lengths even if you seem to say I only know one length how can I use one line to find two lengths you can use one length to find two related legs and so I would ask you to try the same thing for the second question where the idea is to make this right angle triangles appear yourself and use symmetry to help you a circle is always symmetric so you get some Symmetry and I&#39;m not just talking about aob and cob being symmetric I can even just add in a line of symmetry for this triangle and that would give me a new triangle with length 7 and a right angle so I can call this height H and the radius r do the same thing here so that you get a rectangle this is seven this is Hitch and this little bit here is the radius minus seven and so you will have two equations seven hit and R as well as the tiny one over there hitch R minus seven and six from here gas and track is not entirely wrong but the easiest way is to just subtract the two equations and then solve for R right that&#39;s what you&#39;re supposed to find now you notice that we have been making algebra just pop up everywhere algebra and geometry questions algebra in our so-called numerical questions and you may be a little bit sick of algebra here&#39;s something that is pretty legitimately not algebra divisibility tests and remainder something that you have seen before in Primary School but I can assure you that there are still harder questions that can be asked than the primary school Olympians an example would be the first one we see over here to use the divisibility test there are two issues one there is no divisibility test for six but that&#39;s quite easily resolved six is two times three so just do a test for two and three now two that straightforward the number is odd but for three how do I know the remainder when it is divided by 3 the test for three sum up the digits and then divide by three to get the remainder oh that sounds awful how am I supposed to do that well we can upgrade our divisibility test a little bit and so the upgrade that we can do is that the divisibility test does not just work exactly as it is told to you for example for the test for three or nine you can upgrade it to the sum of just contiguous segments which sounds like an awfully fancy word but what I mean is that you can also split it up anywhere any way you like and add up those things instead of adding it up digit by digit so in order to get the remainder when dividing by three I can do a simpler task which is to add up from 1 to 2021 and when I add up from 1 to 2021 it&#39;s not really with the intention of adding up from 1 to 2021 I just want the remainder when that&#39;s divided by three so I know how to add up numbers from 1 to 2021 it would be 2021 numbers times the average which is 20 22 over 2. before I calculate this 1011 is already a multiple of three so therefore the remainder when divided by 3 is just 0. but that&#39;s not the answer is it because we are not saying that this number is divisible by 6. it is divisible by 3 and it is odd so what can the remainder be when it is divisible by 3 but odd it can&#39;t be one because if the remainder is 1 that means that it was not divisible by 3. the only possibility is that it is 3. so know your divisibility tests well and for the second question the same procedure split it into 3 and 11 and for 11. your test is going to be an alternating sum but make sure you know where to alternate from if you are trying to find a remainder let&#39;s say if you wanted to find the remainder when 4321 is divided by 11. well if you just did direct division you get 392 remainder 9. so if you said alternating sum and say okay I do 4 minus 3 plus 2 minus 1 is 2 wait that&#39;s not right and indeed it is not because for the alternating sum for 11 it needs to be from right to left in other words 1 minus two plus three minus four which gives you negative two and negative 2 means that it is two less than a multiple of 11. and so that&#39;s how you get 9 as your remainder so make sure you know your test and these are not just tests for divisibility these are also tests for what the remainder is equal to as I mentioned another comment it&#39;s worthwhile to think about why these are true but I do call this a last minute Smo crash course right so I think now is not the time to think about why it is true now is the time to just make sure that you can apply it correctly another word on number Theory this time highest common factor and lowest common multiples if you&#39;re like me my first introduction to the highest common factor in primary school is using this table which my teacher called like a hdb block if you want to find let&#39;s say the highest common factor of something like 20 and 30 you say okay I take out a tool and then after that I take out a tool again no I can&#39;t take out a tour I can take out a five and after that you stop and say that okay there&#39;s nothing left already and so my hcf based on the left column is 2 times 5 is 10. and this is probably how your thoughts the first time and this is perfectly good if you know what the numbers are isn&#39;t it we don&#39;t know what the numbers are for a question like this so I can&#39;t even try to draw this hdb block out I want you to have this notion of every number being like a list of Demands every number is like a list of Demands and that List of Demands is based on its prime factorization so here you can guess but I&#39;m just going to show you the full method which is that let&#39;s say for 80 and 40 you can prime factorize it as 2 cubed times five and 2 to the power of 4 times 5 12 is 2 squared times 3 and 252 that will take a little while it&#39;s 2 squared times 3 squared times seven so each number is a list of Demands and also you can think of it as a list of resources so 80 contains four tools and a five highest common factor is asking you what resources do these two numbers both possess and so this here tells you that X possesses a 2 cubed and a five it does not have a 2 to the power of 4 otherwise that would have been in common for the second statement 2 squared and 2 squared so it means that well X had two cubed so it&#39;s okay it also has two squared so if it has three tools it also has two tools however it doesn&#39;t have 3 squared it only has one tree otherwise 3 squared would have been in common it also does not have a seven otherwise that would have shown up and so therefore you compare this this it means that the minimum X would need to have two cubed it will need to have a tree and it will need to have a five so that is equal to 120 coincidentally enough it is actually true that 120 is the lowest common multiple of 40 and 12. does it make sense why that&#39;s a good idea well the lowest common multiple is kind of the reverse of that so in the second question the lowest common multiple looks for the bigger one it looks for if let&#39;s say here 42 is 2 times 3 times 7 and 14 is 2 times 7. why are they asking me for 3 well that means that X must have a 3. and you can reason similarly for the second part so the prime factorization may seem like a little bit of over Q here but I can assure you that you would really like to be able to use the prime factorization for many purposes and one of the purposes that is quite interesting is that you can count factors using the prime factorization for instance if I give you a huge number 2 to the power of 7 times 3 to the power of 11 times 5 to the power of four you would be thinking okay how long is it going to take me to calculate this what if I ask you how many factors does it have well there is actually a very simple shortcut and all you need to do is to take the powers and add one to them and multiply together so seven plus one is eight eleven plus one is twelve four plus one is five and so I can say that this has 480 factors why does that work the reason is actually not so complicated if we continue on this discussion of factors meaning looking for resources for a number to be a factor of this 2 to the seven times three to the power of 11 times 5 to the power of four a factor must look like 2 to the power of something 3 to the power of something and five to the power of something you can&#39;t say that my Factor includes the number 19 No 19 is nowhere to be found in that prime factorization we also can say that the factor of 2 cannot exceed 7 because that&#39;s as many as is available if you took 2 to the power of 8 that&#39;s not a factor and so your options in the power are 0 to 7 0 to 11 and 0 to 4 and yes I mean zero zero means don&#39;t use that prime number what&#39;s wrong with that one&#39;s a factor and one would be 2 to the power of 0 3 to the power of zero and five to the power of zero so nothing wrong with that and 0 to 7 means seven plus one choices 0 to 11 means eleven plus one choices zero to four is four plus one choices and then you multiply them to find the total number of options so you can see here that this is a very useful tool and in fact the questions that I&#39;ve selected here are relatively tame by comparison because it just says four factors and three devices which is the same as factors but you are asked to find the numbers or to count the numbers that&#39;s why you can&#39;t just say that I&#39;m just going to list out four factors what&#39;s so hard about that because you are asked the reverse question so it helps to know that this is the way to count factors so that you know four factors could either be three plus one or four is two times two which is one plus one times one plus one what does that mean three plus one corresponds to a prime number cubed one plus one and one plus one means that you have two different prime numbers with the power being one so p q where p and Q are prime numbers and with that we just need to count how many numbers are like that the two examples six is two times three is one of the second type and eight equals to two cubed is one of the first time so we just need to list it out that has to be done carefully of course but it is now a systematic task and not just factoring the numbers one by one by one so P cubed the only thing that fits is 3 cubed which is 27 and everything else goes past 50. for PQ is the product of two prime numbers and two times just about everything is going to fit so 2 times 5 7 11 13 17 19 and 23 are all of the prime numbers that can fit that and then you have a handful more three times five three times seven three times eleven three times thirteen three times Seventeen is 51 so we don&#39;t include that and then 5 times 7 and that is basically the end of the list everything else is going to exceed and we cannot use a prime squared a prime squared would have three factors not four I think I just bought the second question didn&#39;t I so in total here if we count it there are a total of 13 numbers and I&#39;ve included this question just to mention one thing don&#39;t be afraid to do a bit of listing as long as you know what you&#39;re doing listing is not some mathematical sin you are allowed to list as long as your list is complete and your list is not going to be 700 numbers long this thing is okay so not everything is going to be some complicated method or some sophisticated theorem and to conclude some things are just common sense you can&#39;t really put a finger on a theorem if we look at the first question we might be thinking how am I supposed to actually find that two digit number that is 18 more than the product of its two digits well forget about thinking too hard the answer is 99. because 99 works and obviously you can&#39;t find anything larger there are going to be problems like this and problems like these are the thoughts that are challenging those of you who may be a bit too eager to solve the question and then you start cataloging theorem after the theorem of the theorem formula after formula remember these questions are supposed to be fine there are also some others that are just a bit strange looking like the second question if you looked at the second question it doesn&#39;t seem very hard to solve because 10 minus 2x is just negative two times of x minus 5. so if I just cancel this off m is going to be negative 2x minus 1. or rather another another way of thinking about it is that X is 1 plus M over negative 2 wait how do I have no Solutions in x so there&#39;s some questions which are like this and you just need to look back at your steps and sit back and think what could have gone wrong with my steps how do I find m without finding X but how do I find X when they tell me there&#39;s no solution in X so there&#39;s no X isn&#39;t there no M so this is very confusing and very weird well the only thing that can go wrong is when you cancel off x minus 5. the problem is that what if x equals to five well you might say then don&#39;t let it be five precisely so if x equals to 5 is the solution then it is not a solution and so if this was 5 then this is actually going to be not a solution so you want to be able to analyze questions that are unusual that&#39;s at the heart of Olympians and so at the end of the day I&#39;ve just shown you 10 ideas for the assemble Junior but remember just to have some fun and also two reason things out on your own you&#39;re all better at math than you think you are so reason things out don&#39;t just rely on formulas don&#39;t just rely on having seen the question before that&#39;s the entire Spirit of doing math olympiads and on that note I&#39;ll be concluding this mini crash course feel free to comment below if you have any questions and also feel free to comment below if you&#39;d like to discuss any of the second questions here which I didn&#39;t go through in detail so all the best everyone and I hope this helped a little bit

Transcript for:Preparing for SMO Junior Exam Success

Transcript for:
Preparing for SMO Junior Exam Success