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-2B 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-2B 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-2B+7, which is a multiple of seven since, from above, A-2B 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.