What can you say about a number with remainder 1 and 2 when divided by 3 and 4 respectively?

Question

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?

Answer 1

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$.

Answer 2

This is a consequence of the Chinese Remainder Theorem: https://en.wikipedia.org/wiki/Chinese_remainder_theorem

Answer 3

In base-twelve, with digits 0123456789AB,

• Multiples of 3 end in the digit 0, 3, 6, or 9. Numbers that are 1 more than a multiple of 3 thus end in 1, 4, 7, or A (ten).
• Multiples of 4 end in the digit 0, 4, or 8. Numbers that are 2 more than a multiple of 4 thus end in 2, 6, or A.

So in order for a number to meet both criteria, its dozenal representation must end in the digit A. This means it's 2 less than a multiple of twelve.