Let the sum of two three-digit numbers be divisible by 37. Prove that the six-digit number obtained by concatenating the digits of those numbers is also divisible by 37.
$\overline {abc}$ + $\overline {def}$ is divisible by 37. Prove $$\overline{abcdef}$$ is divisible by 37.
$$\overline {abc} = 100a + 10b + c$$
$$\overline {def} = 100d + 10e + f$$
then we have
$$\overline {abc}+ \overline {def} = 100a + 10b + c + 100d + 10e + f = 100(a+d) + 10(b+e) + c + f $$
And I'm stuck here. Can anyone help me?
Best Answer
$$\overline{abcdef}=1000\cdot \overline {abc}+\overline {def}$$ $$=999\overline {abc}+\overline {abc}+\overline {def}$$ $$=(37\cdot 27 \cdot \overline {abc})+(\overline {abc}+\overline {def})$$
Hope this helps.