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Can somebody post the answer in the systematic explained way?? For the sake of new and weak users? I mean how do you start to think when you see this problem. The systematic approach can be developed for this kind of problems?? And by the way what is POE?

q^2 - 5 = 30k + r now if we take r =0 then q can never be a perfect square as 6k+1 is never a multiple of 5.

That is not true, Eg. 25,55,85,.... are all multiples of 5 of the form 6k+1

Proof that \(q^2-5\) cannot be a multiple of 30

q^2 = 30k + 5 = 6*(5k) + 5 So q^2 leaves remainder 5 when divided by 6.

Consider the various cases for the remainder that q leaves when divided by 6 : q=6k ... q^2 will leave remainder 0 when divided by 6 q=6k+1 ... q^2 will leave remainder 1 when divided by 6 q=6k+2 ... q^2 will leave remainder 4 when divided by 6 q=6k+3 ... q^2 will leave remainder 3 when divided by 6 q=6k+4 ... q^2 will leave remainder 2 when divided by 6 q=6k+5 ... q^2 will leave remainder 1 when divided by 6

So it is not possible to have a perfect square that leaves remainder 5 when divided by 6.

So 30k+5 is not a perfect square for any choice of k. _________________

I am sorry but still i am not sure if i face this problem in exam i will be able to crack this.. How you will target at 30?? I mean you have all other options around...??? Is there any way to crack this type of problems??

q^2 - 5 = 30k + r now if we take r =0 then q can never be a perfect square as 6k+1 is never a multiple of 5.

That is not true, Eg. 25,55,85,.... are all multiples of 5 of the form 6k+1

Proof that \(q^2-5\) cannot be a multiple of 30

q^2 = 30k + 5 = 6*(5k) + 5 So q^2 leaves remainder 5 when divided by 6.

Consider the various cases for the remainder that q leaves when divided by 6 : q=6k ... q^2 will leave remainder 0 when divided by 6 q=6k+1 ... q^2 will leave remainder 1 when divided by 6 q=6k+2 ... q^2 will leave remainder 4 when divided by 6 q=6k+3 ... q^2 will leave remainder 3 when divided by 6 q=6k+4 ... q^2 will leave remainder 2 when divided by 6 q=6k+5 ... q^2 will leave remainder 1 when divided by 6

So it is not possible to have a perfect square that leaves remainder 5 when divided by 6.

So 30k+5 is not a perfect square for any choice of k.

This is a good explanation but how do you apply this on the other answers quickly? 30 is easily divided to 5 and 6. but what about 41 or 29? you have to think about a number which these numbers are multiplied by and than add 5 which is a perfect square. Not trivial at all I think.

Any shortcut suggestions? What's the source of the question? _________________

Now q-1, q, q+1 are three consecutive integers , one of them should be a multiple of 3

If q-1 or q+1 is a multiple of 3; q^2-5 is not because 4 is not a multiple of 3 if q is a multiple of 3, q^2 is a multiple of 3 but q^2 -5 is not a multiple of 3

So q^2 -5 is never a multiple of 3, so it will not be a multiple of 30

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...