Equation $a+k \pmod{b+k} = 0$ and the smallest $k$

Question

We let's equation:

$a+k \pmod{b+k} = 0$

For example:

$a = 153$; $b = 52$;

We are looking for the smallest $k$ giving solution

$153+k \pmod {52+k} = 0$

The correct result is $k=49$ becouse:

$153+49 \pmod {52+49} = 202 \pmod {101} = 0$

What is the quickest way to calculate $k$? I need a very quick way.

I have never seen that notation before. Does it mean $a+k\equiv 0 \pmod {(b+k)}$? — lulu, Nov 01 '16 at 14:02

score 0 · Accepted Answer · answered Nov 01 '16 at 14:02

0

a-b (mod b+k) is also 0;

k is the (minimal divisor of a-b greater then b) - b

answered Nov 01 '16 at 14:02

kotomord

Any faster way exist? – Aurelio Nov 01 '16 at 14:10
probably no - is the equivalent of the factorisation problem – kotomord Nov 01 '16 at 14:59

1 Answers1