Congruence with variable modulus

Question

I have been working on some problems and one of them has been particularly challenging. The problem is as follows.

Find a non-trivial (meaning more than 1 digit) positive integer a that satisfies:

$a/10 + 3121(a \mod 10) = 0 \mod a$

Here the division operator is meant to denote integer division so algebraically this is equivalent to:

$(\lfloor a/10\rfloor + 3121(a - 10*\lfloor a/10\rfloor)) - a*\lfloor\lfloor a/10\rfloor + 3121(a - 10*\lfloor a/10\rfloor))/a)\rfloor = 0$

I believe there should be some straightforward way to calculate this value. Using a brute force guess and check I found that $a = 101$ satisfied this congruence but I wanted to know if there was a way to analytically work this one out?

Answer 1

What about $\,a=3\,$? According to what you wrote, and I'm not saying I fully understand how, where, when...of it, we get:

$$\left[\frac{3}{10}\right]+3,121\cdot 3=0+3\cdot 3,121=0\pmod 3$$

And the above , of course, works the same for any positive integer $\,a<10\,$ ...

Added: If you want $\,a\,$ with more than one digit take $\,a=103\,$ :

$$\left[\frac{103}{10}\right]+3,121\cdot (103\pmod{10})=10+3\cdot 3,121=9,373=103\cdot 91$$