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n*n = 72*x = 2*2*2*3*3*x, where x is some random number

If we take the square root of both sides, we have

n = 2*3*sqrt(2x)

We know that n must be an integer so sqrt(2x) must be a positive integer. This means that x must be an odd exponent of 2. The lowest possible value of sqrt(2x) is therefore 2.

All that said, n must be a factor of 2*3*2=12, so (B) should be the correct answer.

1)If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n? a. 2 b. 3 c. 4 d. 6 e. 8

for the first one, suppose p = 4(any prime integer greater than 2, lets suppose 3)= 12

the dividers of 12 are: 1, 2, 3, 4, 6, and 12. altogather 6. so D is the answer.

Priti wrote:

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

for the second one, n^2 is a square of a positive integer and is also divisible by 72, whose factors are 2x2x2x3x3. so n could be any square number that is divisible by 72. for example 144, 144x4, 144x9, 144x16, 144x25, 144x36. remember n^2 cannot be 72 because n is a positive integer and sqrt(n^2) must be an integer.

lets see with some square values:
if n^2 = 144, n = 12 and n is divisible by 1,2,3,4,6, and 12. so 12 is the largest positive integer.
if n^2 = 144x4, n = 24 and n is divisible by 1,2,3,4,6,8, 12 and 24. so 24 is the largest positive integer. but n^2 could be 144 or 144x4. 24 works for the later one but doesnot work for the former where as 12 works for both so 12 or B is the correct answer.