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The last expression above is exactly what it means to add the digits of a together in blocks of length n, starting from the right.

Running total: We now have divisibility tests for 2, 3, 4, 5, 8, 9, 10, 11, 16, 20, 27, 25, 32, 33, 37, 40, 50, 64, 80, 99, 100, 101.

3.3

Take an Alternating Sum of Digits

To see if a is divisible by 11, alternately add and subtract the digits of a starting from the right; this alternating sum and a leave the same remainder when divided by 11. As in the previous section, we can extend this idea to blocks of digits. For instance, a is divisible by 91 if and only if the alternating sum of blocks of 3 digits is divisible by 91. To see if 23210481381 is divisible

by 91 we consider 381

481 + 210

23 = 87. Clearly 87 is not divisible by

91, so neither is 23210481381.

Alternating

Sum

of

Digits

Trick:

Let d be given, and suppose that

10n

1

(mod d) for some n.

Alternately add and subtract the digits of

a in blocks of n starting from the right, and call the result s. Now a and s leave the same remainder upon division by d; in particular, a is divisible by

d if and only if s is divisible by d.

Below are the values of d (2 d 102) for which the alternating sum blocks are at most 4.

d7

11

13

73

77

91

101

block size to add alternately

3

1

3

4

3

3

2

Why it works: Suppose that 10n

1 (mod d), and we are given the

7

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