carcass wrote:
If m is the square of integer n and m is divisible by 98, m must also be divisible by:
I. 28
II. 196
III. 343
(A) I only
(B) II only
(C) I & II only
(D) II & III only
(E) I, II, and III
I would like to know why the answer (see spoiler) and not B
here we go: 98 has factors : 2 7 and 7 so : \(2^1\) and \(7^2\)
But at the same time \(n\) \(=\) \(m^2\) so m is a perfect square so we need
\(2^2\) and \(7^2\) at least28 = \(2^2\) and \(7\)
196 =\(2^2\) and \(7^2\)
343 = \(7^3\)
Please explain i'm struggling a lot the last days with this concept
Given: \(m=n^2\) (not \(n=m^2\) as you've written) and \(m\) is a multiple of \(98=2*7^2\).
Now, since \(m\) is a perfect square than it must have even powers of its primes, thus \(m\) must be divisible by \(2^2*7^2=196\) and
naturally by all the factors of 196, which are: 1, 2, 4, 7, 14,
28, 49, 98 and
196.
Answer: C.
Hope it's clear.