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 = 2x2x2x
3
70 is divisible by
5 <--> 70 = 2x
5x7
330 is divisible by
6 <--> 330 =
2x
3x5x11
56 is divisible by
8 <--> 56 =
2x
2x
2x7
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 =
2x
2x
2x?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 =
3x
3x?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
