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If m and n are positive integers...

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If m and n are positive integers...

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If m and n are positive integers, is$$\sqrt{n-m}$$an integer?

(1) n>m+15
(2) n=m(m+1)

The OA is B.

Can any expert explain this PS question please? I don't understand. Thanks.

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LUANDATO wrote:
If m and n are positive integers, is$$\sqrt{n-m}$$an integer?

(1) n>m+15
(2) n=m(m+1)

The OA is B.

Can any expert explain this PS question please? I don't understand. Thanks.
S1: Pick some easy numbers
If m = 1, then n > 16
Case 1: m =1 and n = 17. √ (17-1) = √ 16 = 4; YES, we have an integer
Case 2: m =1 and n = 18. √ (18-1) = √ 17, NO, we don't have an integer. S1 alone is not sufficient.

S2: n = m^2 + m. so √ (n - m) = √( m^2 + m - m) = √ (m^2) = m; because we're told that m is an integer, we know that the answer to the question is YES, so S2 alone is sufficient. The answer is B

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