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17 Jan 2023, 09:00
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D
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Difficulty:
95%
(hard)
Question Stats:
33% (02:49) correct
67%
(02:48)
wrong
based on 163
sessions
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Not Attempted Yet
If n is a prime number greater than 65, then which of the following statements must be true?
I. 2n + 1 leaves a remainder of 3 when divided by 4
II. (n-1)(n+7) is divisible by 12
III. (n^2-1)^2 is divisible by 72
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
E. None out of I, II and III
I. 2n + 1 leaves a remainder of 3 when divided by 4
II. (n-1)(n+7) is divisible by 12
III. (n^2-1)^2 is divisible by 72
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
E. None out of I, II and III
18 Jan 2023, 07:53
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Bunuel
Must be (D) all, check by plugging in any Prime number > 65 , ie say 67
20 Aug 2023, 17:59
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n is a prime greater than 65. then n is odd. so n = 2k+1 for k>33
2n+1 = 2(2k+1)+1 = 4k + 3 ..thus it gives a remainder of 3 when divided by 4 => I is always true
(n-1)(n+7) = 2k*(2k+8) = 4*k*(k+4). for this to be divisible by 12, k*(k+4) must be divisible by 3.
now 3 cases : k can be 3m, 3m+1, 3m+2 (m >= 0). when k=3m or 3m+2, it becomes divisible by 3.
for k=3m+1, we cant say for sure if its divisible by 3. however, we need to see if we can get a prime number n with k=3m+1. n = 2k+1 = (6m+2) + 1 = 6m+3. this is NOT a prime number. so k can never take the form 3m+1. for the rest of the forms it can take, 4*k*(k+2) is indeed divisible by 12. therefore II is always true.
(n^2 - 1)^2 = [(n+1)(n-1)]^2 = [2k(2k+2)]^2 = 16*(k^2)*((k+1)^2). for this to be divisible by 72, the k part must be divisible by 9. since both k parts are being squared, it will be sufficient for us to check if either of k or k+1 is a multiple of 3. when k = 3m , we get k^2 to be multiple of 9. for k = 3m+2, we get (k+1)^2 to be multiple of 9 . now k=3m+! case remains. But we saw from above that for k = 3m+1, n cant even be prime. so therefore III must always be true
therefore , answer is D
2n+1 = 2(2k+1)+1 = 4k + 3 ..thus it gives a remainder of 3 when divided by 4 => I is always true
(n-1)(n+7) = 2k*(2k+8) = 4*k*(k+4). for this to be divisible by 12, k*(k+4) must be divisible by 3.
now 3 cases : k can be 3m, 3m+1, 3m+2 (m >= 0). when k=3m or 3m+2, it becomes divisible by 3.
for k=3m+1, we cant say for sure if its divisible by 3. however, we need to see if we can get a prime number n with k=3m+1. n = 2k+1 = (6m+2) + 1 = 6m+3. this is NOT a prime number. so k can never take the form 3m+1. for the rest of the forms it can take, 4*k*(k+2) is indeed divisible by 12. therefore II is always true.
(n^2 - 1)^2 = [(n+1)(n-1)]^2 = [2k(2k+2)]^2 = 16*(k^2)*((k+1)^2). for this to be divisible by 72, the k part must be divisible by 9. since both k parts are being squared, it will be sufficient for us to check if either of k or k+1 is a multiple of 3. when k = 3m , we get k^2 to be multiple of 9. for k = 3m+2, we get (k+1)^2 to be multiple of 9 . now k=3m+! case remains. But we saw from above that for k = 3m+1, n cant even be prime. so therefore III must always be true
therefore , answer is D
