If n is a prime number greater than 65, then which of the following

Question

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

Abhishek009 · Answer

Must be (D) all, check by plugging in any Prime number > 65 , ie say 67

vedha0 · Answer

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 =  ^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

dhruvilsang · Answer

Every prime number greater than 3 is of the form 6k + 1 or 6k - 1.

just assume the above two cases for n for each option and you will get the answer.

xiaxyl · Answer

It works for option II and III but in I if we take 6k-1 then how do we get the result?

Bunuel · Answer

If the prime is of the form 6k - 1, then 2n + 1 = 2(6k - 1) + 1 = 12k - 1. The first term, 12k, is divisible by 4, and -1 divided by 4 leaves a remainder of 3.

If the prime is of the form 6k + 1, then 2n + 1 = 2(6k + 1) + 1 = 12k + 3. The first term, 12k, is divisible by 4, and 3 divided by 4 leaves a remainder of 3.

Thus, option I must be true.

P.S. The process for finding the remainder when dividing a negative integer by a positive integer follows the same principles as when dividing a positive integer by a positive integer.

For example:

What is the remainder when dividing 21 by 6? We find the closest multiple of 6 that is  less  than 21, which is 18. Then, we calculate (dividend) - (closest multiple  less  than the dividend) = 21 - 18, yielding a remainder of 3. 
 What is the remainder when dividing -23 by 7? Here, we find the closest multiple of 7 that is  less  than -23, which is -28. Then, we calculate (dividend) - (closest multiple  less  than the dividend) = -23 - (-28), resulting in a remainder of 5.

What about dividing -100 by 30? The closest multiple of 30  less  than -100 is -120. So, the remainder is (dividend) - (closest multiple  less  than the dividend) = -100 - (-120) = 20.

Alternatively, consider this method:

What is the remainder when dividing -23 by 7? Dividing 23 by 7 gives a remainder of 2. To find the remainder for -23 divided by 7, subtract this 2 from the divisor. Thus, the remainder when dividing -23 by 7 is 7 - 2 = 5.

Similarly, what is the remainder when dividing -100 by 30? Dividing 100 by 30 gives a remainder of 10. Therefore, the remainder when dividing -100 by 30 is 30 - 10 = 20.

maniknct · Answer

by plugging 101 we can see all of the three condition are true. So, Ans: D

Vibhatu · Answer

This is very simple, we have to plug-in numbers. all of them are true. We can try 67, 101 etc.
Answer: D

If n is a prime number greater than 65, then which of the following

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If n is a prime number greater than 65, then which of the following

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