Problem Solving

AAPL wrote:Manhattan Prep

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 and II only
D. II and III only
E. I, II, and III

OA C

Let's factorize 98. 98 = 2*7^2. Since m is a perfect square, it must be divisible by even powers of prime factors 2 and 7. Thus, m must be divisible by 2^2*7^2 = 196; thus m must be divisible by all the factors of 196, i.e., 1, 2, 4, 7, 14, 28, 49, 98 and 196.

The correct answer: C

Hope this helps!

-Jay
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AAPL wrote:Manhattan Prep

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 and II only
D. II and III only
E. I, II, and III

OA C

Since 98 = 2 x 49 = 2 x 7^2 and m is a perfect square, then m must be divisible by (at least) 2^2 and 7^2. In other words, m is divisible by 2^2 x 7^2 = 4 x 49 = 196 and any factors of 196. Since 28 is a factor of 196, then m is also divisible by 28. Thus, m is divisible by 28 and 196. If we take m = 196, we will see that m does not need to be divisible by 343.

Answer: C