The number of positive integers which divide 2004 to leave a remainder of 24

What is the number of positive integers that, when 2004 is divided by them, leave a remainder of 24?

A. 036 B. 020 C. 022 D. 021 E. 014

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What have you tried? –  Qiaochu Yuan Feb 4 '11 at 17:52
i tried using factorization method. i can find the numbers with remainder 0 like this..suppose i have 72.i can go in this way 1*72=72 (remainder=0); 2*36=72 (remainder=0)..and so on....but i cant find the sol. for the above question...:( –  user6625 Feb 4 '11 at 18:08
Look for the divisors of 2004-24. –  lhf Feb 4 '11 at 19:09

Let $a$ be the positive integer you're looking for. We know that $$2004 \equiv 24 \mod{a}$$ So we know, that there is an integer, let's say $k$ for which $$k \cdot a + 24 = 2004$$ Subtracting 24 from both sides yields
$$1980 \equiv 0 \mod{a}$$ From $$1980 = 2^2 \cdot 3^2 \cdot 5 \cdot 11$$ we can tell that $1980$ has $$3 \cdot 3 \cdot 2 \cdot 2 = 36$$ divisors, but we only need the ones that are greater than 24, otherwise there couldn't remain 24 when we divide 2004 with $a$.
The divisors lesser than 25 are: 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 18, 20, and 22; 14 divisors total. So the answer is $36-14 = 22$, that's option C.
@TylerHilton: this has $36$ divisors, since a number of the form $n = p_1^{\alpha_1} \dots p_k^{\alpha_k}$ has $(\alpha_1 + 1) (\alpha_2 + 1) \dots (\alpha_k + 1)$ divisors. –  JavaMan Dec 5 '11 at 20:19