What are the divisibility rules for 56?

Answer 1

56 = 7 × 8, thus the number must be divisible by both 7 and 8 - check for each of these digits:

8 is easy to check: add the units digit to twice the tens digit to 4 times the hundreds digit; if this sum is divisible by 8 then so is the original number.

There is no easy check for 7. One I can offer:

Split the number into blocks of 3 digits starting at the right hand end.

Now alternatively subtract and add the each of the digits from the blocks: the units, tens and hundreds digits separately.

Finally add the sum of the units digits to 3 times the sum of the tens digits to twice the sum of the hundreds digits; if this sum is divisible by 7, then so is the original number.

eg 123456789 → 123 456 789

→ sum units: 9 - 6 + 3 = 6

→ sum tens: 8 - 5 + 2 = 5

→ sum hundreds: 7 - 4 + 1 = 4

→ check sum: 6 + 3×5 + 2×4 = 29

29 is not divisible by 7, so 123456789 is not divisible by 7.

[The remainder of 29 divided by 7 is 1, so 123456789 divided by 7 has a remainder of 1.]

eg is 135792648 divisible by 56?

8 + 2×4+4×6 = 40 which is divisible by 8, so 135792648 is divisible by 8.

135792648 → 135 792 648

→ sum units: 8 - 2 + 5 = 11

→ sum 10s: 4 - 9 + 3 = -2

→ sum 100s: 6 - 7 + 1 = 0

→ chk: 11 + 3×-2 + 2×0 = 5 not divisible by 7, so 135792648 is not divisible by 7

Thus 135792648 is not divisible by 56

eg is 63592648 divisible by 56?

Divisible by 8 as before: 8 + 2×4 + 4×6 = 40 = 5×8

63 592 648

→ sum units: 8 - 2 + 3 = 9

→ sum 10s: 4 - 9 + 6 = 1

→ sum 100s: 6 - 5 = 1

→ chk: 9 + 3×1 + 2×1 = 14 = 2×7 → divisible by 7

→ 63592648 is divisible by 56

(63592648 = 113558×56)