What is the divisibility rule for 7 and 8?

Answer 1

they don't have one.

* * * * *

There is one for 8:

The number formed by the last three digits have to be divisible by 8.

There is an alternative for the 8 rule:

Add the last (units) digit to twice the previous (tens) digit to 4 times the previous (hundreds) digit; if this sum is divisible by 8, so is the original number.

eg is 16737456856352 divisible by 8?

(2) + (2 x 5) + (4 x 3) = 24 = 3 x 8 ⇒ 16737456856352 is divisible by 8.

There are various rules for 7, but they can take just as long, if not longer, than just trying to divide the number by 7.

One rule for 7:

Split the number up into blocks of 3 digits from the right hand end (just like splitting a number up into blocks of 3 digits for reading), eg 12234 would be split as 12 234).

Now for each block create the sum of adding the last digit to three times the middle digit to twice the first digit.

Starting at the right hand end, alternately subtract and add the block sums found.

The remainder of this result when divided by 7 is the same as the remainder when the original number is divided by 7.

eg 12234 → 12 234

→ (2 + 3 x 1 + 2 x 0), (4 + 3 x 3 + 2 x 2)

→ 5, 17

→ 17 - 5 = 12 which has remainder 5 when divided by 7, thus 12234 has a remainder of 5 when divided by 7.

eg 4193645654756754 → 4 193 645 654 756 754

→ (4 + 3 x 0 + 2 x 0), (3 + 3 x 9 + 2 x 1), (5 + 3 x 4 + 2 x 6), (4 + 3 x 5 + 2 x 6), (6 + 3 x 5 + 2 x 7), (4 + 3 x 5 + 2 x 7)

→ 4, 32, 29, 31, 35, 33

→ 33 - 35 + 31 - 29 + 32 - 4 = 28 = 4 x 7, thus 4193645654756754 is divisible by 7.