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Problem 1028

National olympiad, first round

Number theory Difficulty 6.0 Find the answer

4. An infinite geometric progression consists of natural numbers. It turned out that the product of the first four terms equals param1. Find the number of such progressions.

The infinite geometric progression consists of positive integers. It turned out that the product of the first four terms equals param1. Find the number of such progressions.

| param1 | Answer |
| :---: | :---: |
| $2^{200} 3^{300}$ | 442 |
| $2^{200} 5^{400}$ | 578 |
| $3^{200} 5^{600}$ | 867 |
| $2^{300} 7^{600}$ | 1326 |
| $3^{300} 7^{500}$ | 1092 |

## Condition

An infinite geometric progression consists of natural numbers. It turned out that the product of the first four terms equals $2^{200} 3^{300}$ . Find the number of such progressions.

## Answer.

442

Your answer

Official solution

Solution

If $b_{1}$ is the first term of the progression and $q$ is its common ratio, then the product of the first four terms of the progression is $b_{1}^{4} q^{6}$ . Therefore, $b_{1}^{2} q^{3}=2^{100} \cdot 3^{150}$ . Hence, $b_{1}=2^{a} 3^{b}, q=2^{c} 3^{d}$ , and we obtain the system: $2 a+3 c=100, 2 b+3 d=150$ . The first equation has 17 solutions in non-negative integers ( $c$ is an even number, i.e., $c=2 n$ , from which $a+3 n=50$ , and we get $0 \leq n \leq 16$ ), the second has 26 solutions ( $b=3 m, 2 m+d=50$ , i.e., $0 \leq m \leq 25$ ).

Each solution to the first equation can correspond to any solution of the second. The total number of progressions is $17 \cdot 26=442$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.