Which of the following is equal to the least positive integer that is divisible by each of the integers 4 through 10, inclusive?
(A) (2^3)(3^2)(5)(7)
(B) (2^3)(3^3)(5^2)(7)
(C) (2^4)(3^3)(5)(7)
(D) (2^5)(3^2)(5)(7)
(E) (2^6)(3^3)(5^2)(7)
[spoiler]I am surprised how many of PS problems, all relate to Prime numbers/factorization!
Answer: A
To find the least common multiple of several numbers, start by finding the prime factorization of each number:
4 = 2^2
5 = 5
6 = 2 x 3
7 = 7
8 = 2^3
9 = 3^2
10 = 2 x 5
The least common multiple will contain every prime factor, so there must be a 2, a 3, a 5, and a 7 in the result. To find the exponent, look for the largest exponent among the factorizations. For instance, there is a 2 in the factorizations of 4, 6, 8, and 10. The largest exponent is 2^3, the factorization of 8. Thus, the least common multiple will contain 2^3. That limits our choices to (A) and (B).
The largest exponent of 3 is 3^2, in 9. The only exponents of 5 and 7 are 1, so the answer we're looking for is (A), (2^3)(3^2)(5)(7).v[/spoiler]
(A) (2^3)(3^2)(5)(7)
(B) (2^3)(3^3)(5^2)(7)
(C) (2^4)(3^3)(5)(7)
(D) (2^5)(3^2)(5)(7)
(E) (2^6)(3^3)(5^2)(7)
[spoiler]I am surprised how many of PS problems, all relate to Prime numbers/factorization!
Answer: A
To find the least common multiple of several numbers, start by finding the prime factorization of each number:
4 = 2^2
5 = 5
6 = 2 x 3
7 = 7
8 = 2^3
9 = 3^2
10 = 2 x 5
The least common multiple will contain every prime factor, so there must be a 2, a 3, a 5, and a 7 in the result. To find the exponent, look for the largest exponent among the factorizations. For instance, there is a 2 in the factorizations of 4, 6, 8, and 10. The largest exponent is 2^3, the factorization of 8. Thus, the least common multiple will contain 2^3. That limits our choices to (A) and (B).
The largest exponent of 3 is 3^2, in 9. The only exponents of 5 and 7 are 1, so the answer we're looking for is (A), (2^3)(3^2)(5)(7).v[/spoiler]
