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02 Aug 2020, 12:11
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
53% (02:19) correct
47%
(01:51)
wrong
based on 126
sessions
History
Date
Time
Result
Not Attempted Yet
If m and n are prime numbers such that m<n, what is the value of m?
(1) \(m^2+ n^2\) + 2n = 199
(2) n = 13
(1) \(m^2+ n^2\) + 2n = 199
(2) n = 13
03 Aug 2020, 11:09
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Quote:
Step 1: Understanding statement 1 alone
Sum of the terms \(m^2, n^2\) and 2n is 199 ie odd.
Rearranging the terms;
\(m^2 + n^2\) = 199 - 2n
As 2n is always even, therefore \(m^2 + n^2\) = odd - even = odd
As sum of square of two prime numbers is odd, therefore one of them must odd and other be even.
Only even prime number is 2 and it is given that m<n, hence m=2
Sufficient
Quote:
Step 2: Understanding statement 2 alone
Value of m cannot be determined
Insufficient
A is correct
11 Aug 2024, 04:06
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