In what follows I shall be using representation of a number in various base systems. In the
decimal (base 10) system a number a is written as
Theorem
Let b be an integer. Then an integer a
- is divisible by (b-1) if and only if the sum of digits in its expansion (a)b is
divisible by (b-1)
- is divisible by (b+1) if and only if the alternating sum of digits in its expansion (a)b is
divisible by (b+1)
Proof
The proof is based on Modulo Arithmetic. Thus we have
b-1=0 (mod (b-1)) which implies b=1 (mod (b-1)). Therefore, for every i>0, bi = 1 (mod(b-1)). Multiplying this by ci and summing up for i=0,1, ..., k we get the first assertion.
Similarly, b+1=0 (mod (b+1)) hence b=-1 (mod (b+1)). Therefore, for i>0, bi = (-1)i (mod(b+1)).
Multiplying this by ci and summing up for i=0,1, ..., k we get the second assertion.
Example 1
Let a=164. Is it divisible by 4? On the one hand, 164=125+25+2·5+4=(1124)5. The sum of digits is 1+1+2+4=8
and is divisible by 4. Therefore 164 is divisible by 4.
On the other hand, 164=2·81+0·27+0·9+0·3+2=(20002)3. The alternating sum of digits is 2-0+0-0+2=4 and
is divisible by 4. Therefore, again, 164 is divisible by 4.

For divisibility by 7 there are two criteria:
- a = (a)6 = ck6k+ck-16k-1+...+c161+c0 is divisible
by 7 iff the alternating sum of digits (-1)kck+(-1)k-1ck-1+...+c0 is divisible by 7.
- a = (a)8=ck8k+ck-18k-1+...+c181+c0 is divisible
by 7 iff the sum of digits ck+ck-1+...+c0 is divisible by 7.
Example 2
512=83. Therefore (512)10=(1000)8. This implies that divided by 7 the remainder
will be 1. Then, for example, 511 is divisible by 7.
1296=64. Therefore (1296)10=(10000)6. From here (1296+6)10=(10010)6. Its alternating
sum of digits is 1-0+0-1+0=0. As a result, 1302 is divisible by 7.
On Internet
- Divisibility Rules

Copyright © 1996-2009 Alexander Bogomolny