 Hi, I'm Zor. Welcome to Unisor education. So we continue course called mess for mess plus and problems presented on unisor.com This course requires the prerequisite knowledge, which is presented in another course which is called mess for teens on the same website every single the website is free no advertisement and There are some other courses like physics for teens and relativity Okay, so now the course mess plus and problems is dedicated to Unusual problems, which are not really illustration of the theory presented in the previous course mess for teens But rather forces you to think about certain things outside of the box creatively Analytically and come up with certain solutions, which are not obvious at least from the first glance Okay, so today's problems, which we will solve there are four of them They are about numbers and that's why I classify these problems as arithmetic numbers and this is within the arithmetic classification Lecture number six so arithmetic to zero six. All right, so let's solve problems problem number one We have some natural number positive Integer number and which contains only Once and zeros something like this The total number is unknown But I know that number the digit one appears 111 times Number digit zero is a known number of times whatever so Question is Can this number be a square of another integer number? Now as usually I'm encouraging you to pause the video and Think about this problem yourself It's good. It's great. If you come up with a solution If not, just you know continue listening The lecture by the way every lecture on Unusual.com has notes written notes They're just presented parallel to the video so there is a video and there is no And notes contains the same problems in written Format and there are even some hints Sometimes there are solutions as well if it's more difficult problem But sometimes only hints or maybe not not at all more than hints but in any case it makes sense to take a look at the problem in notes on this website and Basically, it's presented the same thing the same way as I presented here and maybe hint will help you Solve it yourself because the best thing if you can solve it yourself That's the whole purpose of the course to basically teach you how to think creatively analytically Unusually So can it be a square? Okay, let's just think about it oops First of all the answer is no it cannot be a square and here is why Now we have 111 times number one now recall that If the sum of the digits of the number is divisible by three Then the number is divisible by three if Some of the digits divisible by nine Then the number itself is divisible by nine. It's very easy to prove. You just have to Present the number as in its decimal notation as something like this This is ten to the power of zero which is one now here if you subtract one from each one minus one minus one minus one and Then add the same thing 10 to the power of zero minus one So basically you subtract a Zero a one these are digits and then add them now these guys are 9999999 Whatever number of nines is so this is divisible by three and by nine That's why this is divisible by three and by nine So if the sum of the digits is divisible by three or nine then the number is divisible by three or nine Now this sum of the digits is a hundred and eleven times one is a hundred and eleven 111 is divisible by three But it's not divisible by nine Now question is if the number is divisible by three, but not divisible by nine Can it be a square? Obviously not because if this number n has certain number of prime numbers which is in its representation Then n square would have two of each I know that there is number three among the this divisors so Every prime number should be at least twice so if the number is divisible by three then it's square divisible by nine and Obviously this cannot be a square because it's divisible by three, but not divisible by nine Some of these some of these digits Okay, that's the first problem now the second problem Okay, there is a number n and There is a number k times n Now what's known about k is that k minus one is not Divisible by three that's a condition. It's a known fact Now what's also known is that the sum of digits of n is Exactly the same as sum of digits of k times n now we have to prove That n is divisible by nine Okay, how can it be proved now? back to Visibility by by three and by nine now the Remainder Again, if you will represent Our number as in this format if you represent our number in this format now this spot Is divisible by three and nine this one might or might not be divisible by three and nine But what I would like to say is that the remainder of Division of the total number By three or nine is exactly the same as remainder of division of some of its digits By three or nine because this piece is divisible by three or nine So if there is a remainder, it's only because this is not divisible by three or nine. Okay like for example take 17 Now 17 divisible by let's say three 15 is the visible remainder will be two some of its digits is eight one plus seven and The result of division of eight by by by three is what? What's the remainder remainder is two because six is the visible so eight has a remainder to the same remainder and The same remainder remainder is because some of the digits is Basically part of the representation of the number in this particular case. Okay, so some of the digits of the number and The number itself have the same remainder dividing by nine now I Know the sum of the digits of this guy is equal to some of the digits of this guy Okay, which means this and this Have the same remainder by division of Divided by nine and this in this Now what does it mean it means that n can be represented as nine times Let's say a one plus some remainder and K times n can also be represented as nine times A times B plus the same remainder Since some of these digits is the same as some of these digits and remainder of division of number and some of its digit is the same So all of these will have the same remainder, which means their difference K and minus n would be divisible B minus a Would cancel so the difference is divisible by nine, but what is this difference? It's K minus one times n K minus one not divisible by three as I said So the only if the whole thing is divisible by nine and this piece is not divisible even by three So this is divisible by nine and that's the answer to this problem Again, we are basing their the base of this Solution is that the remainder In division by nine between number and some of its digits is the same So this and this give the same remainder and These are equal. So this is also the same remainder and if it's the same remainder The difference would be divisible and since K minus one is not divisible by three Then by nine should be divisible. The n should be divisible by nine. Okay. That's my second problem now the third one and Again, don't forget that you are washing this lecture first before you read the notes Don't forget to pause the lecture before I start presenting a solution So the third one Okay, so we have a number nine Which has Only nine As a digit in its decimal representation now Let's assume that the number K Is prime I Actually Not really use this very much, but just for definitiveness Let's just assume that the number of these nines is prime like five for example nine nine nine nine nine Okay, so let's give it Next we have to find another number m which contains only once P times Which is divisible by this one Okay, so again, I would like to find the number Which contains only once Digit one which is divisible by this one Okay Solution I mean post the post the video think about it yourself. Here's the solution Now n can be represented as nine times one one one K times right? That's obvious So if m is divisible by n M should be divisible by nine Which means that the sum of digits of M Should be divisible by nine But it contains only Digits one so some of its digits is basically the number of digits one which is as I said P So basically P should be divisible by nine Because the sum of these digits is exactly as Many digits which is which is P Okay, so first of all we are looking for number which is divisible by nine and That's requiring that the number of digits should be divisible by nine on another side It should be divisible by this number one one one one one P times what are the numbers which are at the same time divisible by this number and Divisible and in the number of these should be divisible by nine Well, let's just take the number which is exactly the same as M So if I will put M that will be one one one one one one it's divisible by this one obviously But I don't know how many digits. I mean I do know that the number of digits is Is is K So basically It's a prime number So it's a prime number so Definitely the this number of digits Is not divisible by nine But what I can do I can put another number in front of it the same another group of This would be K This would be K. So it's 2k. Well, is 2k divisible by nine? No, but if I will continue this and I put nine times Groups of K the number of digits would be what K times nine Since number of digits is K times nine it's divisible by Some of them since all of them are once they will be divisible by by nine and at the same time each group contains K Digits exactly as this one as this one is in its representation as nine times Yes here as this one and each one is divisible. So basically the whole number is this number and Times ten to some power whatever the power is Then another and Plus to again tend to some power Etc, which means it will be divisible by By this by this can Well, not actually and I meant only this part of this and this is and with Bar of the top, okay So we repeat this piece K times digit one nine times in a row and that gives us the visibility by both nine and This and with a bar on the top and that's the answer. That's the number That's the number M, which we are looking for Okay, so this is this Now the force now we have a number and Which is equal to K plus one K plus two Etc to K The product of number numbers where K is Something doesn't really matter Now the question is if I will represent this number and as a product of Prime numbers How many number two will occur? So and it's supposed to be equal to two to some kind of a power Times three to some power et cetera prime numbers question is how many tools will be there So if I will start dividing and by two how many times it will be able to divide and by two Okay Pause the video and here is my answer Now as you see this is the product of all the numbers from K plus one to K Now what if I will add To this product all the numbers from one to K So we'll have one times two times three et cetera times K times K plus one Plus K plus two et cetera to K minus one to K and divide By the same one two three K that would be the same, right? No problem there great now from this from the numerator It has all the numbers from one to two K There are odd numbers and there are even numbers. So let's just separate them one three Five et cetera to K minus one That's odd numbers and even numbers two four six et cetera Okay, and still divide by one two three four five Okay, that's the same thing, right? It's equal Equal to well, let's separate this one three five et cetera to K minus one This is the one part and Look at this. These are Even numbers two four six to K These are each one of them is half of the one which is on the top One is half of two two is half of four three is half of six K is half of two K so we have basically Two times one times two times two Times two times three times two times four Etc. Two times K Divided by one two three etc. Okay, right and these guys are Canceling out and that will have only two How many of twos well obviously K So this is equal to two to the power of K times one three five et cetera To K minus one now. These are all odd numbers. So they do not contain number two among Devisors so all the twos are here. So this number and is Two to the power of K times something something being this So the number of twos is K so you can divide it by by two K times Exactly, and that's the answer to my fourth problem Well, okay, that's it I do suggest you to read the notes for this lecture So you go to unisor.com choose the course math plus and problems then go to the classification of the problems arithmetic and this is zero six in that category That's it. Thank you very much and good luck