Difference between revisions of "2004 AIME II Problems/Problem 8"

Latest revision as of 01:49, 31 July 2025

Problem

How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers?

Solution 1

The prime factorization of 2004 is $2^2\cdot 3\cdot 167$ . Thus the prime factorization of $2004^{2004}$ is $2^{4008}\cdot 3^{2004}\cdot 167^{2004}$ .

We can count the number of divisors of a number by multiplying together one more than each of the exponents of the prime factors in its prime factorization. For example, the number of divisors of $2004=2^2\cdot 3^1\cdot 167^1$ is $(2+1)(1+1)(1+1)=12$ .

A positive integer divisor of $2004^{2004}$ will be of the form $2^a\cdot 3^b\cdot 167^c$ . Thus we need to find how many $(a,b,c)$ satisfy

$(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167.$

We can think of this as partitioning the exponents to $a+1,$ $b+1,$ and $c+1$ . So let's partition the 2's first. There are two 2's so this is equivalent to partitioning two items in three containers. We can do this in ${4 \choose 2} = 6$ ways. We can partition the 3 in three ways and likewise we can partition the 167 in three ways. So we have $6\cdot 3\cdot 3 = \boxed{54}$ as our answer.

Solution 2 (bash)

Clearly we need to find a group of numbers that multiply to 2004. We can list them all out since we know that 2004 is only $167 \cdot 2^2 \cdot 3$ .

167, 2, 2, 3

4, 3, 167

12, 167

4, 501

2, 1002

2, 3, 334

2, 2, 501*

6, 2, 167

3, 668

6, 334

2004*

To begin, the first multiple doesn't work because there are only 3 prime divisors of 2004. We can apply all multiples because the prime factorization of $2004^{2004}$ is $2^{4008} \cdot 3^{2004} \cdot 167^{2004}.$ Every multiple has six ways of distributing numbers to become powers of 167, 3, and 2, except for the ones with a star. For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to. For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1.

Therefore we have $8 \cdot 6$ normal multiples and $3 \cdot 2$ "half" multiples. Sum to get $\boxed{54}$ as our answer.

Solution 3

we want $167^a \cdot 3^b \cdot 2^c$ such that this number has exactly $2004$ positive factors. Note that $2004$ has $(1 + 1) \cdot (1 + 1) \cdot (2 + 1) = 12$ factors.

Case 1: $(a + 1)(b + 1)(c + 1) = 2004$ , and no variable equals $0.$

The three variables can either be the three elements of $\{166, 1, 5\}, \{500, 1, 1\}, \{333, 2, 1\},$ or $\{166, 3, 2\}.$ They can be permuted in $6 + 3 + 6 + 6 = 21$ ways.

Case 2: $(x + 1)(y + 1) = 2004$ , where $x$ and $y$ are two variables chosen out of $\{a, b, c\}$ and none of them equal $0$

There are $12 - 2 = 10$ ways to choose the values of $x$ and $y$ . There are $\binom{3}{2} = 3$ ways to choose the variables $x$ and $y$ out of $\{a, b, c\}$ , giving $30$ ways.

Case 3: $(x + 1) = 2004$

$x$ must be $2003$ . We can let $x = a,b,$ or $c$ , so this case gives us $3$ ways.

The answer is $21 + 30 + 3 = \boxed{54}.$

~grogg007

@@ Line 43: / Line 43: @@
 Therefore we have <math>8 \cdot 6</math> normal multiples and <math>3 \cdot 2</math> "half" multiples. Sum to get <math>\boxed{54}</math> as our answer.
+==Solution 3==
+we want <math>167^a \cdot 3^b \cdot 2^c</math> such that this number has exactly <math>2004</math> positive factors. Note that <math>2004</math> has <math>(1 + 1) \cdot (1 + 1) \cdot (2 + 1) = 12</math> factors.
+'''Case 1:''' <math>(a + 1)(b + 1)(c + 1) = 2004</math>, and no variable equals <math>0.</math>
+The three variables can either be the three elements of <math>\{166, 1, 5\}, \{500, 1, 1\}, \{333, 2, 1\},</math> or <math>\{166, 3, 2\}.</math> They can be permuted in <math>6 + 3 + 6 + 6 = 21</math> ways.
+'''Case 2: ''' <math>(x + 1)(y + 1) = 2004</math>, where <math>x</math> and <math>y</math> are two variables chosen out of <math>\{a, b, c\}</math> and none of them equal <math>0</math>
+There are <math>12 - 2 = 10</math> ways to choose the values of <math>x</math> and <math>y</math>. There are <math>\binom{3}{2} = 3</math> ways to choose the variables <math>x</math> and <math>y</math> out of  <math>\{a, b, c\}</math>, giving <math>30</math> ways.
+'''Case 3:''' <math>(x + 1) = 2004</math>
+<math>x</math> must be <math>2003</math>. We can let <math>x = a,b,</math> or <math>c</math>, so this case gives us <math>3</math> ways.
+The answer is <math>21 + 30 + 3 = \boxed{54}.</math>
+~[[User:grogg007|grogg007]]
 == See also ==

2004 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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