I'm not sure if my proof is correct:



$7|(2a+4b)$ => $7|2(a+2b)$

I'm not sure if my notation is universal but $abb$ symbolizes three digit numbers who's two last digits are the same.

  • $\begingroup$ To finish you need to justify why $\ 7\mid 2n\,\Rightarrow\, 7\mid n\ \ $ $\endgroup$ – Bill Dubuque Apr 6 '17 at 16:08

Using your notation:

$$7|abb \iff 100a+10b+b=7k,\quad k\in\mathbb{N}, \quad 1\leq a,b\leq9$$

We need to prove that $$2a+b=7k,\quad k\in\mathbb{N}$$

We have that:

$$100a+11b=7k\implies 98a+2a+4b+7b = 7k$$

$$7\cdot(14a+b)+2a+4b=7k\implies 2\cdot(a+2b)=7k', \quad k'=k-(14a+b)$$

Since $7\not | 2$, we have that $7|a+2b$ as required.

Edit: Upon re-reading your question, yes your proof does seem correct. Treat the above as an alternate method.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.