Carcass wrote:
If an integer is divisible by both 27 and 10, then the integer also must be divisible by which of the following?
A. 4
B. 25
C. 36
D. 54
E. 81
-----ASIDE---------------------
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, then k is "hiding" within the prime factorization of NConsider these examples:
24 is divisible by
3 because 24 = (2)(2)(2)
(3)Likewise, 70 is divisible by
5 because 70 = (2)
(5)(7)
And 112 is divisible by
8 because 112 = (2)
(2)(2)(2)(7)
And 630 is divisible by
15 because 630 = (2)(3)
(3)(5)(7)
-----ONTO THE QUESTION!---------------------
Let's say the integer in question is N The integer (N) is divisible by 2727 =
(3)(3)(3)So, we can say N =
(3)(3)(3)(?)(?)...
Aside: the other ?'s represents other possible prime numbers in the prime factorization of NThe integer (N) is divisible by 1010 =
(2)(5)So, we can say N =
(2)(5)(?)(?)...
When we combine both pieces of information we can write: N =
(3)(3)(3)(2)(5)(?)(?)
The integer (N) also must be divisible by which of the following? N =
(3)(3)(3)(2)(5)(?)(?)
A. 4
4 = (2)(2)
Since the prime factorization of N (above) does NOT include two 2's, it need not be the case that N is divisible by 4
Eliminate A
B. 25
25 = (5)(5)
Since the prime factorization of N (above) does NOT include two 5's, it need not be the case that N is divisible by 25
Eliminate B
C. 36
36 = (2)(2)(3)(3)
Since the prime factorization of N (above) does NOT include two 2's and two 3's, it need not be the case that N is divisible by 36
Eliminate C
D. 54
54 = (2)(3)(3)(3)
Since the prime factorization of N (above) DOES include one 2 and three 3's, we can be certain that N is divisible by 54
Answer: D