DIVISIBILITY RULE FOR 7

A whole number, N, is a multiple of 7 if the following procedure leads to
another multiple of 7*:
(1) Subtract the ones digit from N,
(2) Dividing the result by 10, and
(3) Subtract—from that result—twice the original ones digit.
*the multiple of 7 may be 0 or negative.
This is best explained in an example. Consider 7 x 39 = 273, so 273 is a multiple of 7. Applying the
Divisibility Rule for 7 to 273 yields the following:

(1) Subtract the ones digit from N: Subtract 3 from 273: 273
– 3
270
(2) Dividing the result by 10: Divide 270 by 10: 270 ÷ 10 = 27
(3) Subtract—from that result—twice
the original ones digit: Subtract, from 27, twice 3:

27

– 6

21:  a multiple of 7
Since this procedure leads to a multiple of 7, it must be that the original number, 273, is a
multiple of 7, as shown in this long division:
A much quicker way to use the procedure is shown here
2 7 3
– 6
2 1
x 2
Since 21 is a multiple of 7, it must
be that 273 is a multiple of 7.

Let’s apply this more direct procedure to a larger multiple of 7: 7 x 568 = 3,976:
3 9 7 6
– 1 2
3 8 5
x 2
We don’t know if 385 is a multiple of 7
or not, so we continue the procedure.
3 8 5
– 1 0
2 8

Since 28 is a multiple of 7, it must be that
both 385 and 3,976 are multiples of 7.
What is the proof behind this procedure?
Consider, without loss of generality, a three digit number, N = 100a + 10b + c, where a, b, and c are
single digits, a ≠ 0.
Applying the Divisibility Rule for 7 to 100a + 10b + c, we get:
(1) Subtract c: 100a + 10b
(2) Divide by 10: 10a + b
(3) Subtract twice c: 10a + b – 2c
Claim: 100a + 10b + c is a multiple of 7 if and only if 10a + b – 2c is a multiple of 7
P

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