If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A) 10 B) 11 C) 12 D) 13 E) 14
Thanks!
Integer Question
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Easy, first factor the numberDFritschy wrote:If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A) 10 B) 11 C) 12 D) 13 E) 14
Thanks!
990= 2*3^2*5*11
So if the product is the multiple of 990, then n must be greater than or at least 11.
The answer is B.
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DFritschy wrote:If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A) 10 B) 11 C) 12 D) 13 E) 14
Thanks!
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 N
Similarly, we can say:
If N is is a multiple of k, 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
So, if if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.
Since 990 = (2)(3)(3)(5)(11), we know that a 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.
For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.
So, the answer is B
Cheers,
Brent