if n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is a.6 b.12 c.24 d.36 e.48

sanaa.rizwan wrote:if n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is a.6 b.12 c.24 d.36 e.48 n² must be divisible by 72. Since n² is the square of an integer, the value of n² must be a perfect square: 1, 4, 9, 16, 25, 36, 64, 81, 100, 121, 144... In the list above, the smallest value divisible by 72 is 144. If n² = 144, then n = 12. This is the MINIMUM value of n. Since n=12 is not divisible by 24, 36, or 48, eliminate C, D and E. Since n cannot be any smaller than 12, it must be divisible by answer choice B (12). The correct answer is B.

sanaa.rizwan wrote:if n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is Let us assume, n² = 72k = (2³)*(3²)*k, where k is some positive integer. We can see that, minimum possible value of k such that (2³)*(3²)*k becomes a square of an integer is 2. Hence, minimum possible value of n² is (2�)*(3²). So, minimum possible value of n is (2²)*3 = 12. The correct answer is B.

n^2 / 72 n x n / 2x2x2x3x3,so each n must have 2x2x3 = 12 (minimum value on n) so ans 12

sanaa.rizwan wrote:if n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is a.6 b.12 c.24 d.36 e.48 From the question, How do we know that we have to find the MINIMUM value of n ? Thank you !

gughanbose wrote:From the question, How do we know that we have to find the MINIMUM value of n ? Because we can always make n sufficiently large such that all the options (even numbers larger than the provided options) will divide n. For example, if n = 72*48, then all the given options divided n. But, there will be some values of n, for which 24 or 36 or 48 will not divide n. Hence, the idea is to make n as small as possible. The factors of the smallest n will always divide other larger values of n. Hope that helps.