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If M and N are integers such that M = N^2 – 1, is M divisible by 12?
(1) N - 1 is the square of an even number
(2) N + 1 is a two digit number and the sum of its digits is a multiple of 3
(1) N - 1 is the square of an even number
(2) N + 1 is a two digit number and the sum of its digits is a multiple of 3
06 Feb 2022, 08:16
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M= N^2-1
=> M= (N+1)*(N-1)
Either N+1 or N-1 should be divisible by 12
or
N+1 and or N-1 should have 3 & 4 as divisors
St 1) N-1 is a square of an even #
=> N-1 is divisible by 4
Picking numbers for N-1, N, N+1
N-1 = 4 (square of 2) [4,5,6]
N-1 = 16 (square of 4) [16,17,18]
=> (N-1) * (N+1) are divisible by 12 ...therefore sufficient
St 2) N+1 is a two digit # and sum of digits is multiple of 3
=> N+1 is divisible by 3
N+1 could be 12,36,48 ..sufficient
N+1 15,18 ...not sufficient
Therefore only St1) sufficient...(A)
=> M= (N+1)*(N-1)
Either N+1 or N-1 should be divisible by 12
or
N+1 and or N-1 should have 3 & 4 as divisors
St 1) N-1 is a square of an even #
=> N-1 is divisible by 4
Picking numbers for N-1, N, N+1
N-1 = 4 (square of 2) [4,5,6]
N-1 = 16 (square of 4) [16,17,18]
=> (N-1) * (N+1) are divisible by 12 ...therefore sufficient
St 2) N+1 is a two digit # and sum of digits is multiple of 3
=> N+1 is divisible by 3
N+1 could be 12,36,48 ..sufficient
N+1 15,18 ...not sufficient
Therefore only St1) sufficient...(A)
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06 Feb 2022, 13:47
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Bunuel
Bunuel
If M and N are integers such that M = N^2 – 1, is M divisible by 12?
(1) N - 1 is the square of an even number
(2) N + 1 is a two digit number and the sum of its digits is a multiple of 3
(1) N - 1 is the square of an even number
(2) N + 1 is a two digit number and the sum of its digits is a multiple of 3
Bumping!!! Knockout this one and get KUDOS for a correct solution!!!
Statement 1:
N-1 = (2p)²
N = 4p²+1
M = (4p²+1)²-1
M = 16p⁴+1+8p²-1
M = 16p⁴+8p²
M = 8p²(2p²+1)
p can be 3k, 3k+1 or 3k+2/3k-1
M = 8(3k)²(2*9k²+1) = 72k²(18k²+1)
>> Divisible by 12
M = 8(3k+1)²(2*(3k+1)²+1) = 8(9k²+1+6k)(18k²+3+12k) = 24(9k²+6k+1)(6k²+4k+1)
>> Divisible by 12
M = 8(3k-1)²(2*(3k-1)²+1) = 8(9k²+1-6k)(18k²+3-12k) = 24(9k²+1-6k)(6k²+1-4k)
>> Divisible by 12
Therefore, M = 8p²(2p²+1) will always be divisible by 12. Hence, statement 1 alone is sufficient.
Statement 2:
N+1 is a 2-digit number, and the sum of its digits is a multiple of 3
Case1: Let N+1=98, therefore N=98
M = 98² - 1
M will have an odd value
Can't be a multiple of 12
Case2: Let N+1=96, therefore N=95
M = 95² - 1
M = 96*94
M will be a multiple of 12
Therefore, statement 2 alone cannot give us a conclusive answer
Hence, only statement 1 alone is sufficient, option A.
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