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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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    Look for the divisors of 2004-24.2011-02-04

1 Answers 1

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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.

(And here's the full list of solutions (I used brute force and ruby): $30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 165, 180, 198, 220, 330, 396, 495, 660, 990, 1980$.)

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    @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.2011-12-05