What can you say about a number with remainder 1 and 2 when divided by 3 and 4 respectively? I was trying to solve a problem which states:


  How many two-digit numbers have remainder 1 when divided by 3 and remainder 2 when divided by 4?


and solved it by writing down individual numbers...

Looking at the solution, I was baffled by the explanation :


  The lowest common multiple of 3 and 4 is 12. Hence both of the required
  conditions are satisfied only by numbers that are 2 less than multiples of 12
  and also less than 100, ie: 10, 22, 34, 46, 58, 70, 82 and 94.


Why on earth does the number have to be 2 less than multiples of 12? Could someone offer a better explanation please?
remainder 1 when divided by 3
$x \equiv 1 \pmod 3$
$x \equiv -2 \pmod 3$

remainder 2 when divided by 4
$x \equiv 2 \pmod 4$
$x \equiv -2 \pmod 4$

And so
$x \equiv -2 \pmod 3$
$x \equiv -2 \pmod 4$

Since 3 and 4 are relatively prime, this implies that we can combine the two equations to
$x \equiv -2 \pmod {12}$

Another way of looking at this is

$x \equiv -2 \pmod 3$ means that $x = 3n - 2$ for some $n$, or $x + 2 = 3n$ for some n.

Similarly, $x + 2 = 4m$ for some m.

Hence $3n = x + 2 = 4m$.
Since $\gcd(3,4) = 1, $ then m must be a multiple of 3.

It follow that $x + 2 = 4m = 12r$ for some $ m = 3r.$

finally, $ x = 12r - 2$.
This is a consequence of the Chinese Remainder Theorem: https://en.wikipedia.org/wiki/Chinese_remainder_theorem