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Re: If x is the number of positive integers that will leave no remainder [#permalink]
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Answer would be E.

To find the number of different factors of a number, we conduct prime factorization, add '1' to each of the powers of primes, and then multiply the powers.

So 1880 = 2^3*5^1*47^1. Hence 4*2*2 = 16 (E)
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If x is the number of positive integers that will leave no remainder [#permalink]
Bunuel wrote:
If x is the number of positive integers that will leave no remainder when divided into 1,880, what is the value of x?

A. 4
B. 6
C. 8
D. 12
E. 16


This problems test us on the number of prime factors of the number.

\(1880\)= \(2^3\) x \(5^1\) x \(47^1\)

Total factors = ( 3+1 ) ( 1 + 1 ) ( 1 + 1 ) = 16

Hence answer is 16

PS : There is nothing important in this problem except understanding the highlighted part, if you get the meaning correctly it will be a 1 minute job
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Re: If x is the number of positive integers that will leave no remainder [#permalink]
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Re: If x is the number of positive integers that will leave no remainder [#permalink]
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