Go forth and multiply

Number Theory Level pending

Let $n\leq 10^6$ be a $k$ digit positive integer written as $n=\overline{d_1d_2\cdots d_k}$ .

Let $n_1,n_2,\cdots,n_p$ be the $p$ solutions for the equation

$n=k^2\prod_{i=1}^{k}d_i$

What is the value of $\sum_{i=1}^{p} n_i$

The answer is 324.

1 solution

Janardhanan Sivaramakrishnan
Aug 13, 2015

All single digit integers i.e., $1,2,3,4,5,6,7,8,9$ are solutions.

In case of 2-digit numbers, we need $10d_1+d_2=4d_1d_2$ . Since it is a 2-digit number $d_1 > 0$

Hence, we can modify the equation as $10+\frac{d_2}{d_1}=4d_2$ .

Thus, we need $4d_2 > 10$ and $d_1$ being a factor of $d_2$ and $10+\frac{d_2}{d_1}$ is a multiple of 4.

For the first condition, $4d_2 > 10$ , we have choice of $d_2=3,4,5,6,7,8,9$

Obviously, $d_2 \ne d_1$ (11 is not a multiple of 4). Hence, $d_2$ is composite.

This reduces the choice to $d_2=4,6,8,9$

Making $4d_2 = 16,24,32,36$ . But, none of the choices lead to integral values for $d_1$ . Hence, there are no 2-digit solutions.

The logic has already become so convoluted that I would no longer pretend to know an analytical way to find more solutions. I would resort to statements.

There are two 3-digit solutions $135$ and $144$ .

There are no further solutions less than $10^6$ .

Thus, the required answer is

$1+2+3+4+5+6+7+8+9+135+144=\boxed{324}$

Go forth and multiply

The answer is 324.

1 solution

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