If n = t^3 for some positive integer

ritumaheshwari02 wrote:If n = t^3 for some positive integer t and if 8, 9 and 10 are each factors of n, which of the following must be a factor of n?

A. 16
B. 81
C. 175
D. 225
E. 275

A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k (i.e., k is a factor of N), then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7

Okay, onto the question.

We're told that 8 is a factor of n, which means 8 must be hiding in the prime factorization of n.
In other words, we know that n = 2x2x2x?x?x?... (the ?'s represent other values that could also be in the prime factorization of n)
IMPORTANT: Since n = (t)(t)(t), we can conclude that 2 must be a factor of t.

Similarly, told that 9 is a factor of n, which means 9 must be hiding in the prime factorization of n.
In other words, we know that n = 3x3x?x?x?...
IMPORTANT: Since n = (t)(t)(t), we can conclude that 3 must be a factor of t.

Using similar logic, we can conclude that 5 must be a factor of t.

So, if 2, 3 and 5 are all factors of t, we know that t = 2x3x5x?x?x?...

Since n = t^3, we know that n = (2x3x5x?x?x?)(2x3x5x?x?x?)(2x3x5x?x?x?)

In other words, n = 2x2x2x3x3x3x5x5x5x?x?x?x?x?x?...

From here, it's easy to see what must be a factor of n.

A) Since we can only be certain that n has three 2's hiding in its prime factorization, we cannot conclude that 16 is a factor of n

B) Since we can only be certain that n has three 3's hiding in its prime factorization, we cannot conclude that 81 is a factor of n

C) 175 = 5x5x7. Since we 7 may or may not be hiding in the prime factorization of n, we cannot conclude that 175 is a factor of n

D) 225 = 3x3x5x5. Since there are two 3's and two 5's hiding in the prime factorization of n, we can conclude that 225 is a factor of n

E) 275 = 5x5x11. Since we 11 may or may not be hiding in the prime factorization of n, we cannot conclude that 275 is a factor of n

The correct answer is D

Cheers,
Brent

I think 700-level is about right.

Cheers,
Brent