If n is a positive integer and n squared is divisible by 72, then the largest possible integer that must divide n is
a. 6
b. 12
c. 24
d. 36
e. 48
a. 6
b. 12
c. 24
d. 36
e. 48
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Picking Number Approachshriti wrote:If n is a positive integer and n squared is divisible by 72, then the largest possible integer that must divide n is
a. 6
b. 12
c. 24
d. 36
e. 48
The prime factor of 72 are -> 2, 2, 2, 3, 3shriti wrote:If n is a positive integer and n squared is divisible by 72, then the largest possible integer that must divide n is
a. 6
b. 12
c. 24
d. 36
e. 48
n>0 (n is integer)shriti wrote:If n is a positive integer and n squared is divisible by 72, then the largest possible integer that must divide n is
a. 6
b. 12
c. 24
d. 36
e. 48
Anurag@Gurome wrote:Picking Number Approachshriti wrote:If n is a positive integer and n squared is divisible by 72, then the largest possible integer that must divide n is
a. 6
b. 12
c. 24
d. 36
e. 48Algebraic Approach:
- Least possible value of n² such that n² is divisible by 72 is 72*2 = 144
Hence, minimum possible value of n = 12.
Largest possible integer that divides n is 12.The correct answer is B.
- n² is divisible by 72
Hence we can write n² as 72k, where k is an positive integer.
Now, n = √n² = √(72k) = √[(2)*(36)*k] = 6√(2k)
Now for n to be an integer, k must be an even multiple of a perfect square.
Hence, we can write k = 2m², where is a positive integer.
Now, n = 6√(2k) = 6√(2*2*m²) = 12m
Hence, largest possible integer that divides n is 12
The trick here is "MUST", because the smallest possible value of n² is 144 (2² * 2² * 3²) the largest possible divisor of n MUST be 12. In this case 12 is not divisible by 48.tomada wrote:Please, why is 48 not the answer? 48^2 is divisible by 72, and 48 divides 'n'.
By the way, what is the source of this question?
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