 Hello friends, so welcome again to this problem solving session again. This session is To solve a problem related to Euclid's division lemma and application of Euclid's division lemma now the question says and We have seen that how to approach a question first try to gather all the information after reading the question And then we'll see what and what should the approach of solving that question So prove that this is the question so prove that what one of every three three consecutive positive integers, this is the information three consecutive positive integers is Divisible by three so this is what they are Asking us to prove so let us take an example to understand so every example of three consecutive positive integers. So Six seven eight these are three consecutive positive integers and then say they are saying One of these so one of these is divisible by three indeed Indeed one of three one of these three is divisible by three. This one is divisible by three. So this is three into two So hence guess looks like it's true. Okay, so let's take another example. So another example could be 11 12 13 Okay, so this is example one Example one and this is example two So let's say we are now we are now just trying to understand whether whatever is given is actually valid or not So example two is 11 12 30 and clearly this number 12 is divisible by three So let's say 20 example three example three you can take example three example three could be that's a 27 28 29 yes, this is divisible by three another example for example could be one not one One not two one not three. So again, this number is divisible by Three another could be so, you know, there are there could be a infinite number of such examples So let's say one not again three one not four and one not five So if you see this number again is divisible by three So every set of three consecutive positive integers one is coming out to be multiple of three Okay, so how do we generalize this? Observation so how do we start? So actually this is the key if you see this three this three is the hint that we can start Expressing any number in the form of any positive integer again in the form of what all 3k 3k plus one and 3k plus two Isn't it so any any number can we express the 3k 3k plus one 3k plus two any positive integer and any Let's say you any three consecutive numbers will be so three three three consecutive Conjective consecutive positive positive integers integers are of The form what are of the form What kind of a form so they will be if the first is n second will be n plus one and n plus two Is it it so first is n second will be n plus one How it will be n plus two you can also take it you can take it as n minus one and And n plus one this is also again. You can give three numbers if n is where n is n is a positive Positive integer. So if you if you take it like that if you take it like that Then there are multiple ways you can express. So let us take this expression So n n plus one n plus two and now let us say case one case one Let's say n is n can be of what all form n can be 3k form This form n could be this form and n could be 3k plus two only three possibilities of Expressing n into n as a you know multiple of three and some remainder. So n let's say first case is n is 3k form n is of 3k form, okay So if you see if n is 3k then our problem is solved anyways, why because the three numbers will be so three numbers three consecutive consecutive integers integers would be what integers would be 3k then 3k plus one and 3k plus two clearly out of these three three k is Multiple so this one is multiple of multi multiple of three So case one is valid. So case one case one holds true Holds true Let us take the other two cases. What are the other two cases? Let's say let's say if case two What is case two? Case two is when n is of form 3k Plus one so first case is n is 3k second is n is equal to 3k plus one Then how what what will happen? So the three consecutive again three consecutive consecutive Integers would be would be what all? 3k plus one and then the next number will be one added to it that is 3k Plus two and third one will be 3k Plus three So clearly if you see the third number this one so this number the third number this can be expressed as three times k Plus one hence hence a multiple of three So again case two also holds So one of these three Case two holds true. Okay for case two also. This is This is great. Now take us. Let us take case three. What is case three? case three when n is is equal to 3k Plus two is the third case one case is 3k second case is 3k plus one and third case is 3k plus two Isn't it now if n is 3k plus two then three consecutive Conjective consecutive means Continuous successive consecutive integers Integers Would be would be what? 3k plus two Then second one will be 3k plus three and third one will be 3k plus four Out of which if you see if you this if you take This this number this one this is this is of the form of The form Three m or 3k right because it can be written as three times K plus one hence we say that that So hence hence a multiple of hence a multiple of Three okay, so hence we see in all the three cases hence this also case case three Also holds true holds true. So in all the three cases we see so so hence hence hence in all the cases in all the three cases In all three cases one of one of three three consecutive consecutive Integers positive consecutive positive integers positive integers is a multiple multiple of Three hence hence proved hence proved So hope you understood the question so or the solution and This was what just a recap. So question was proved that one of every three consecutive positive integers is observed by three That is is multiple of three So we take two three cases when n is 3k 3k plus one and 3k plus two and in all the cases be evaluated three consecutive integers and in in One of the three cases was always of the form or way as always divisible by three hence proved Thanks for watching this video We will be having more number of problem solving sessions in future. Thanks a lot