maniac wrote:
If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) m – n = 3
Solution:
Given that p, m and n are positive integers. If n is odd, n^2 must also be odd. How do you represent an odd integer? As (2k + 1)
So n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1
Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. Therefore, when n^2 is divided by 8, it will leave a remainder of 1.
Statement 1: When p is divided by 8, the remainder is 5.
When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5. When n^2 is divided by 8, the remainder will be 1. To get a remainder of 5, when m^2 is divided by 8, we should get a remainder of 4.
m^2 = 8a + 4 (i.e. we can make ‘a’ groups of 8 and 4 will be leftover)
m^2 = 4(2a + 1)
This implies m = 2*√(Odd Number) because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that m is not divisible by 4.
This statement alone is sufficient.
Statement 2: m – n = 3
The difference between m and n is 3 i.e. an odd number. Since n is odd, we can say that m will be even (Even – Odd = Odd). But whether m is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p.
This statement alone is not sufficient.
Answer (A)
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