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?

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    $\begingroup$ $999 \equiv 0 \mod 37$ $\endgroup$ – ParaH2 Dec 26 '15 at 14:16
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    $\begingroup$ If you should face down a problem about $37$ again: The observation that $3 \times 37 = 111$ is usually relevant (as it was/is here). $\endgroup$ – Benjamin Dickman Dec 27 '15 at 5:03

$$\overline{abcdef}=1000\cdot \overline {abc}+\overline {def}$$ $$=999\overline {abc}+\overline {abc}+\overline {def}$$ $$=(37\cdot 27 \cdot \overline {abc})+(\overline {abc}+\overline {def})$$

Hope this helps.


Let $x$ denote $\overline{abc}$.

Let $y$ denote $\overline{def}$.

Hence $1000x+y=\overline{abcdef}$.

Then $\color\red{x+y=37n}\implies1000x+y=999x+\color\red{x+y}=999x+\color\red{37n}=37(27x+n)$.


Notice, since $\overline {abc}+\overline{def}$ is divisible by $37$ hence, $$(100a+10b+c)+(100d+10e+f)=37\lambda $$ or $$100d+10e+f=37\lambda-(100a+10b+c)\tag 1$$ where, $\lambda$ is some integer

Now, one should have concatenated number as $$\color{blue}{\overline{abcdef}}=100000a+10000b+1000c+100d+10e+f$$ $$=1000(100a+10b+c)+100d+10e+f$$ setting value from (1), $$=1000(100a+10b+c)+37\lambda-(100a+10b+c)$$ $$=999(100a+10b+c)+37\lambda$$ $$=37(2700a+270b+27c+\lambda)$$ since, $(2700a+270b+27c+\lambda)$ is an integer, $37(2700a+270b+27c+\lambda)$ is divisible by $37$ i.e. $\color{red}{\overline{abcdef}}$ is divisible by $\color{red}{37}$


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