Problem Solving

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

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

The prime factor of 72 are -> 2, 2, 2, 3, 3

n² is divisible by 72 HENCE 2, 2, 2, 3, 3, ...? (2², 2, 3²)are prime factors of n² ("...?" because there could be more prime factors)

Because n² must have at least (2², 2, 3²) in its prime factors, and n² = n * n, we can distribute this factors to determine the prime factors of n:

n²
=
n -> 2 * 3 * 2
*
n -> 2 * 3 * 2 We must fill this space with "2" to obtain two equal values of n

[spoiler]So the largest number that must divide n is 2 * 3 * 2, which is 12 (1, 2, 3, and 6 could divide n but 12 is the largest one)

Correct Answer is ...B...[/spoiler]

Thanks Anurag and Maaj.. your explanations was really helpful..

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?

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

n>0 (n is integer)
N^2/72=i (i is integer)
prime factorization of 72
72--2--36
36--2--18
18--2--9
9---3--3
3---3--1
72=2^3 * 3^2; to make 72 the perfect square we multiply by 2^1 --> 2^4 * 3^2=(4*3)^2
12^2 is divisible by 72; n=12

@Tomada: 72=2^3 * 3^2; to make 72 the perfect square we multiply by 2^1 and 3^2 --> 2^4 * 3^4=(4*9)^2
36^2 is divisible by 72; n=36

@Tomada: 72=2^3 * 3^2; to make 72 the perfect square we multiply by 2^3 --> 2^6 * 3^2=(8*3)^2
24^2 is divisible by 72; n=24
...

so basically we are missing one important statement in this problem:

If n is a positive integer and n squared is divisible by 72, then the largest possible integer that must divide the lowest value of n is

a. 6
b. 12
c. 24
d. 36
e. 48

the correct answer is B then

@Tomada -->

Anurag@Gurome wrote:

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

Picking Number 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.

Algebraic Approach:

• 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 correct answer is B.

but how is it implicit that you must find the minimum value of 'n'?

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?

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.

It could be that n² is greater than 144 but we are not sure about that!!! in that case n could be divisible by 48

I think this problem is from Manhattan GMAT number properties.

Thanks Maaj - just to build on what you said earlier, I think we have to take the whole phrase in itself..."the largest possible integer that MUST divide n is"...because of "must" it's like saying "the lowest possible integer that could divide n is"...is that correct? Is this a common pattern found in gmats?