How many ordered integer solutions $(a,b,c)$ does the equation $abc = 1000$ have?

The answer is 400.

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Well presented! I loved how you broke up the complex problem into three simple parts of distributing the powers of 2, 5 and the signs, and then used the rule of product to get the final result.

Pranshu Gaba
- 4 years, 3 months ago

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I loved the observation that the total positive solutions is the sum of number of divisors of each divisor of $1000$ , which is in fact the dirichlet convolution $(1 \star \sigma_0) (1000)$ .

I am thinking how can we use this method if there are four integers $a, b, c, d$ such that $abcd = 1000$ . Continuing along the same lines, for given $a$ and $b$ , we have as many choices for $c$ as there are positive divisors of $\frac{1000}{ab}$ . Once $a, b, c$ are chosen, $d$ is fixed. How can we evaluate the sum $\displaystyle\sum_{a|1000} \sum_{b|\frac{1000}{a}} \sigma_0\big(\tfrac{1000}{ab}\big)$ ?

Pranshu Gaba
- 4 years, 3 months ago

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The key fact is that the constant function $1$ is multiplicative, and that Dirichlet convolutions of multiplicative functions are multiplicative. Thus $\sigma_0 = 1 \star 1$ is multiplicative, for example. But then $\begin{aligned} 1(p^j) & = 1 \\ (1 \star 1)(p^j) \; =\; \sigma_0(p^j) & = j+1 \\ (1 \star 1 \star 1)(p^j) \; = \; (1 \star \sigma_0)(p^j) & = \sum_{k=0}^j \sigma_0(p^k) \; = \; \tfrac12(j+1)(j+2) \\ (1 \star 1 \star 1 \star 1)(p^j) & = \; \sum_{k=0}^j (1 \star 1 \star 1)(p^k) \; =\; \tfrac16(j+1)(j+2)(j+3) \end{aligned}$ and so on, for any prime $p$ and integer $j \ge 0$ . Thus $(1 \star 1 \star 1 \star 1)(1000) \; = \; (1 \star 1 \star 1 \star 1)(2^3) (1 \star 1 \star 1 \star 1)(5^3) \; = \; 20^2 \; = \; 400$ and this last is the sum you want.

Mark Hennings
- 4 years, 3 months ago

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Ah yes, thank you :) This makes it clear. For $n$ integers, the equation $a_1 a_2 a_3 \ldots a_n = 1000$ would have $(\underbrace{1 \star 1 \star 1 \star \cdots \star 1}_{n \text{ times}}) (1000)$ solutions.

$(\underbrace{1 \star 1 \star 1 \star \cdots \star 1}_{n \text{ times}}) (p^j)$ is equal to $\binom{j + n-1}{n -1}$ . We get

$\begin{aligned} (\underbrace{1 \star 1 \star 1 \star \cdots \star 1}_{n \text{ times}}) (1000) &= (1 \star 1 \star 1 \star \cdots \star 1) (2^3) (1 \star 1 \star 1 \star \cdots \star 1) (5^3) \\ & = \binom{3 + n - 1}{n - 1}^2 \\ & = \binom{ n + 2}{3}^2 \end{aligned}$

Pranshu Gaba
- 4 years, 3 months ago

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Since $1000 = 2^3\cdot 5^3$ , each of $a,b,c$ must have the form $\pm 2^m5^n$ for some non-negative integers $m,n$ . As multiplication is commutative and associative, this allows us to break the problem up into three pieces:

Consider just the exponents of $2$ first. If $A,B,C$ are the exponents of $2$ in $a,b,c$ , respectively then we must have $A+B+C = 3$ . Since $A,B,C$ are non-negative integers, we can count the number of ways to choose these exponents using "stars-and-bars". This gives $\binom{3+2}{2} = 10$ ways to determine the exponents of $2$ .

The exponents of $5$ are determined in the same way, giving $\binom{3+2}{2} = 10$ ways to determine the exponents of $5$ .

Then, since the product is positive, there must be an even number of negatives. Therefore, either all are positive ( $1$ way) or exactly two are negative ( $3$ ways). Therefore, there are $1+3 = 4$ ways to choose the signs.

Finally, just multiply: $\binom{3+2}{2} \cdot \binom{3+2}{2} \cdot 4 = \boxed{400}.$