# How many 3 digit numbers give the remainder $8$ when divided by $12$, and $2$ when divided by $10$?

In other words:

Let $$x\in \mathbb N \wedge 100\leq x\leq999$$

$$x\equiv 8 \pmod{12}\equiv2 \pmod{10}$$

I tried to write down the number as $$x=12c+8=10k+2$$ with $$(c,k)\in \mathbb N^2$$ but that didn't help very much.

We consider \begin{align} x &\equiv 8 \pmod{12} \iff \begin{cases} x \equiv 0 \pmod 4 \\ x \equiv 2 \pmod 3 \end{cases} \\ x &\equiv 2 \pmod{10} \iff \begin{cases} x \equiv 0 \pmod 2 \\ x \equiv 2 \pmod 5 \end{cases} \\ \end{align}

which simplifies to

\begin{align} x &\equiv 0 \pmod 4 \\ x &\equiv 2 \pmod 3 \\ x &\equiv 2 \pmod 5 \end{align}

which simplifies to

\begin{align} x &\equiv 0 \pmod 4 \\ x &\equiv 2 \pmod{15} \\ \end{align}

Noting that $(-1)(15) + (4)(4) = 1$, we see that

\begin{align} -15x &\equiv 0 \pmod{60} \\ 16x &\equiv 32 \pmod{60} \\ \end{align}

we conclude that $x \equiv 32 \pmod{60}$ or $x = 32 + 60n \quad (n \in \mathbb Z)$.

Solving $32+60n = 100$ for $n$, we get $n = 1 \frac{2}{15}$. So the first three-digit number that meets our requirements is $x = 32+60(2) = 152$.

Solving $32+60n = 100$ for $n$, we get $n = 16 \frac{7}{60}$. So the last three-digit number that meets our requirements is $x = 32+60(16) = 992$.

From $n=2$ to $n=16$ there are $15$ three-digit numbers that solve $x\equiv 8 \pmod{12}\equiv2 \pmod{10}$.

## A Shortcut

$x \equiv 2 \pmod{10}$ clearly implies that $x = 10n + 2$ for some $n$.

$x \equiv 8 \pmod{12}$ implies

\begin{align} 10n + 2 &\equiv 8 \pmod{12} \\ 5n + 1 &\equiv 4 \pmod 6 \\ 5n &\equiv 3 \pmod 6 \\ -5n &\equiv -3 \pmod 6 \\ n &\equiv 3 \pmod 6 \\ n &= 6t+3 \quad (n \in \mathbb Z) \\ \hline x &= 10(6t+3) + 2 \quad (n \in \mathbb Z) \\ x &= 60t + 32 \quad (n \in \mathbb Z) \\ \end{align}

and we can go on from there as we did before.

HINT

Let $x=12c+8$. Note that this simplies $$12c +8 \equiv 2 \pmod {10}$$ So $$12c \equiv 4 \pmod {10}$$ Dividing each side by $4$, we have $$3c \equiv 1 \pmod 5$$ Multiplying each side by $2$, we have $$6c \equiv c \equiv 2 \pmod {5}$$ So $c$ is of the form $5k+2$. This gives us that $x$ is of the form $60k+32$.

• So $k=2$ is the minimum value and $k=16$ is the maximum value of $x$, thus there are $15$ solutions! Thank you for your time, have a good day. – Glycerius Feb 5 '17 at 12:00