# The Center of Math Blog

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## Monday, July 13, 2015

### Problem of the Week

Good morning from the Center of Math! Here's the latest Problem of the Week, which we'll have up on Facebook, Twitter, Tumblr, and Google+ shortly.

There are a number of ways to go about solving this problem; here's my method. I began by multiplying both sides of the definition of x by 2, then adding a to both sides and subtracting 2b:
Rearranging this equation in two different ways yields:
If x is divisible by 7, then so is 2x, that is, 2x = 7k for some integer k. Then, from equation (5),
a and k are integers, so 3a - k is as well, thus a - 2b is divisible by 7: this is the "only if" in the problem statement. Working the other way,  if a - 2b is divisible by 7, then it equals 7j for some integer j: using equation (4), I found:
So 2x must be divisible by 7, and since 2 isn't divisible by 7, x must be: this is the "if" in the statement, and since we've shown both the "if" and the "only if," we've proven the statement true.

As you might have guessed, this is a handy test for seeing if a base-10 integer is divisible by 7: just subtract twice the final digit from the number formed by the rest of the digits, and if the result is divisible by 7, so is the  original number.

Did you think of a different way to show that this formula works? Let us know in the comments, and be sure to check out the rest of our Problems of the Week.

#### 1 comment:

1. Similar.

Suppose 7|(a-2b).
Then a-2b = 7k, for some integer k,
so a = 7k + 2b,
so 10a + b = 70k + 21b,
and RHS is clearlly divisible by 7.

Suppose 7|x.
Then 10a + b = 0 mod 7
so 3a + b = 0 mod 7
so 9a +3b = 0 mod 7.
But now x = 10a + b = (a - 2b) + (9a + 3b) = 0 mod 7,
and 7 divides the second bracket so must also divide the first.