Apart from

dividing an

integer by

7, there is another method which is often easier for testing whether

7 divides it. Remove the rightmost

digit from the number you're testing, and subtract double this digit from the remaining portion. The result of this

calculation will be divisible by 7

if and only if the original number is

divisible by 7. The chop-double-and-subtract procedure can therefore be repeated until you obtain a number whose

divisibility by 7 can be checked with the naked eye.

For example, here is how the method works in determining whether 86415 is divisible by 7:

- Remove and double 86415's last digit (2 times 5 gives 10). Subtract this from the remaining portion: 8641 - 10 =8631.
- The question has been reduced to determining whether 7 divides 8631; so repeat the procedure using 8631: slice off the trailing 1, double it and subtract from 863 to leave 861.
- One more step: beginning with 861, remove the 1, double it and subtract from 86 to leave 84, which is twelve times seven.
- Since we ended up with a multiple of seven (84) via this procedure, the original number (86415) must also be a multiple of seven.

The advantage of this method over straight division is that less mental calculations are necessary, and those which are required only involve addition and subtraction.

**Proof**

Here's a sketchy outline of half a proof which employs induction. Assume that some multiple of seven (call it *X*) has the property that it remains a multiple of seven after undergoing the chop-double-subtract procedure described above. (We note that 14 satisfies this, since 1 - 8 = -7 is a multiple of 7.) Then we claim that the next highest multiple of seven (*X*+7) also has this property:

Let the last digit of *X* be *B*, and the number formed by removing this digit be *A*. (For example, if *X* is 8631 then *A*=863 and *B*=1.) Then we know that removing *B* from *X* and subtracting it twice from *A* leaves behind a multiple of seven; that is, *A*-2*B* is a multiple of seven. Now consider the next multiple of 7 after *X*, that is, *X*+7. If *B* is 0, 1 or 2, then the rightmost digit of *X*+7 is just *B*+7, so performing chop-double-subtract on *X*+7 will give *A*-2(*B*+7) which is a multiple of 7 since, from above, *A*-2*B* is. On the other hand, if *B* is greater than 2, then the last digit of *X*+7 will be *B*-3 and the leading portion of *X*+7 will be *A*+1. (For example, if *X* is 994 then *X*+7 is 1001: the rightmost digit decreases by 3, while the left portion increases by 1 from 99 to 100.) In this case, performing chop-double-subtract on *X*+7 gives (*A*+1)-2(*B-3*)=*A*-2*B*+7, which is a multiple of seven since, from above, *A*-2*B* is.

The other half of the proof--showing that no *non*-multiple of seven becomes a multiple of seven after having its righmost digit removed and doubly subtracted--is left as an exercise for the interested reader.