Number Theory

**mathisgood** · May 8th 2010, 06:52 AM

Help on this...

The digits of a five-digit number are MNOPQ. Prove that MNOPQ is divisible by 7 if and only of the number MNOP - 2 x Q is divisible by 7.

Thanks!

**Anonymous1** · May 8th 2010, 07:05 AM

Originally Posted by mathisgood

Help on this...

The digits of a five-digit number are MNOPQ. Prove that MNOPQ is divisible by 7 if and only of the number MNOP - 2 x Q is divisible by 7.

Thanks!

Divisibility Tests -- from Wolfram MathWorld

**Soroban** · May 8th 2010, 08:48 AM

Hello, mathisgood!

The digits of a five-digit number are $ABCDE.$
Prove that $ABCDE$ is divisible by 7 if and only of the number $ABCD - 2E$ is divisible by 7.

[1] Given: . $ABCDE$ is divisible by 7.
. . .Prove: . $ABCD-2E$ is divisible by 7.

We have: . $10(ABCD) + E \;=\;7p\:\text{ for some integer }p.$

Subtract $21E\!:\;\;10(ABCD) - 20E \;=\;7p - 21E \;=\;7(p - 3E)$

Divide by 10: . $(ABCD) - 2E \;=\;\frac{7(p-3E)}{10}$

Since $(ABCD)-2E$ is an integer, then $p-3E$ is a multiple of 10: . $10k$

And we have: . $(ABCD) - 2E \;=\;\frac{7(10k)}{10} \;=\;7k$

Therefore: . $(ABCD)-2E$ is divisible by 7.

[2] Given: . $(ABCD)-2E$ is divisible by 7.
. . .Prove: . $ABCDE$ is divisible by 7.

We have: . $(ABCD) - 2E \;=\;7q\,\text{ for some integer }q$

. . . Then: . $(ABCD) \;=\;2E + 7q$

Multiply by 10: . $10(ABCD) \;=\;20E + 7q$

. . Add $E\!:\;\;10(ABCD) + E \;=\;21E + 7q$

And we have: . $ABCDE \;=\;7(3E + q)$

Therefore: . $ABCDE$ is divisible by 7.

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