# Reversible numbers

“Find all of the five-digit numbers that are reversed when multiplied by (a) 9 and (b) 4.”

I can’t say I’m a fan of this type of puzzle, but a student asked if I’d help them prep some number theory, so here we are.

### How I went about it

Well, with a great deal of bad language and worse grace, frankly. But here’s my eventual thinking for (a):

- Let $n = 10^4 a + 10^3 b + 10^2 c + 10 d + e$, such that $N = 10^4 e + 10^3 d + 10^2 c + 10b + a$.
- Considering $9n - N = 0$ gives $89,999a + 990b + 800c - 910d - 9,991e = 0$
- $a = 1$ and $e = 9$, or else $n$ would have six digits.
- So $89,999 + 990b + 800 c- 910d - 89,919 = 0$, or $80 + 990b + 800c + 910d = 0$ - or even $8 + 99b + 80c - 91d = 0$
- Working modulo 10: if $b =0$, $d=8$; if $b=1$, then $d=7$.
- Two options:
- if $b=0$ and $d=8$, then $8 + 80c - 728 = 0$ and $c = 9$
- if $b=1$ and $d=7$, then $8 + 99 + 80c - 637 = 0$ and $80c = -530$, which doesn’t work.

- So $n = 10,989$ and $N = 98,901$.

In a similar vein, I got that $n = 21,978$ and $N = 87,912$ for part (b).

### But hang on a minute!

I noticed a few things here:

- The $n$ for part (b) is double that of part (a)
- In part (a), $n = 11,000 - 11$ and $N = 99,000 - 99$; in part (b), $n = 22,000 - 22$ and $4n = 88,000 - 88$.

And I wondered a few things:

- What about other five-digit numbers of the same form?
- It turns out $(33,000 - 33) \times \frac{7}{3} = 77,000, - 77$, $(44,000 -44)\times\frac{3}{2}$ works, and so does $(55,000 - 55)\times 1$ - although that one is hardly unique.

- What about numbers of different lengths? Is there a pattern?
- Is there significance to the fact that $\frac{1}{91} = 0.\dot 01098\dot9$?

What do you find when you play with it?