   SEARCH HOME Math Central Quandaries & Queries  Question from Rahul: I have to prove that n^2 is a multiple of 100 is necessary or Sufficient condition (or both) for n being multiple of 10. I can see that if n=10*p for some p then n^2 = 100*p^2, and p^2 is also an integer as p is. n^2= 100*p => n=10* sqrt(p) and here sqrt(p) need not be integer. I still feel that n^2 = 100*p will imply n=10*p for two different p's. Also I can not see any immediate counter example to prove that n^2=100*p for p integer, does not imply n = 10*p or q etc. Hi Rahul,

I can give you a hint. $100 = 4 \times 25.$

Write back if you need more assistance.

Penny

Rahul wrote back

I had been given a hint to the problem that 100 = 4*25.
I can say 4*25 = 10*10 which is multiple of 10, what next?

Rahul,

You are assuming that $n^2 = 100 \times p$ and thus $n^2 = 4 \times 25 \times p.$ Therefore $4$ divides the right side and hence it divides the left side. Hence $4 = 2^2$ divides $n^{2}.$ Can you complete the problem now?

Penny      Math Central is supported by the University of Regina and the Imperial Oil Foundation.