This weird algebra number theory question

Question

Find all positive integers $n$ such that:

$n = \prod_{k=0}^{m} (a_k +1)$

where $a_ma_{m-1}... a_1a_0$ is decimal representation of $n$ .

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Now I tried for $m=0$ (Trivially Not possible), $m=1$ : $a_1a_0 = (a_0+1)(a_1+1)$ , hence $a_1(9-a_0) = 1$

Hence $a_1 = 1, a_0 = 8$ . So only possible scenario is $n = 18$ .

Edit:

Trying for $m = 3$ . Simplifying using algebra and then applying parity to eliminate 5 cases for 3 cases trying to use mod 4

I dont know how to proceed.

Pls help. Thanks!!!!

Thanks to @Especially Lime to add the correct tags.
I am new

Answer 1

Notice that $n$ has $m+1$ digits, so $10^m\leq n<10^{m+1}$ . Also, the maximum possible value of $\prod_{k=0}^m(a_k+1)$ is $10^{n+1}$ , if all digits are $9$ . If we have too many digits less than $9$ , then $\prod_{k=0}^m(a_k+1)$ will quickly get too small.

However, suppose we have even one digit $9$ . Then $\prod_{k=0}^m(a_k+1)$ is a multiple of $10$ , so we also have to have at least one digit $0$ . Now the maximum possible value is only $10^m$ , which is too small. (We know $n\neq 10^m$ , since it contains a $9$ .)

Therefore every digit is at most $8$ . But for three digits, all at most $8$ , the maximum value of the product is $729$ , so we can't have three $8$ s. You should be able to continue this argument to show that there are no three-digit solutions (and adapt it to show there are no solutions with more digits).