How to tell if a number is divisible by 7
There are two ways to answer this and neither needs a memorised digit trick. Take away 7s you already know until what is left is small enough to recognise, or chop off the last digit, multiply it by 5 and add it back. Both work on any number, however long.
Take away multiples of 7 you already know until what is left is small enough to recognise from the 7 times table.
How to check it
- Find a multiple of 7 you know that fits inside the number: 7 × 100 = 700, 7 × 600 = 4200, and so on.
- Take it away and look at what is left.
- Repeat until the leftover is small enough to recognise from the 7 times table.
- If you land on 0, the number divides by 7. Anything else is the remainder.
Worked examples
Is 4620 divisible by 7?
4620 − 4200 (that is 7 × 600) = 420 420 = 7 × 60
Nothing left over, so yes. 4620 = 7 × 660.
Is 1147 divisible by 7?
1147 − 700 (7 × 100) = 447 447 − 420 (7 × 60) = 27 27 − 21 (7 × 3) = 6
6 left over, so no. 1147 = 7 × 163 remainder 6.
Chika's trick: chop, multiply by 5, add
In 2019 a twelve-year-old called Chika Ofili was given a holiday maths book and came back with a genuinely new shortcut for 7: chop off the last digit, multiply it by 5, and add it to what is left. Repeat until the number is small enough to judge.
4620 → 462 + 5 × 0 = 462 → 46 + 5 × 2 = 56, and 56 is 7 × 8. So 4620 divides by 7.
It works because 50 is one more than 49, and 49 is seven 7s. Multiplying the last digit by 5 and adding it back does not change whether the number divides by 7. It does change what is left over, so the trick answers yes or no and never gives the remainder: 15 becomes 26, and 26 leaves 5 over where 15 left 1.
The older version of this trick doubles the last digit and subtracts it. That works too, but subtraction asks a child to take a bigger number from a smaller one as soon as the digits are unkind, which is why the adding version is easier to get right.
Why these methods work
Taking away 7s cannot change the answer. Every chunk that comes off is a whole number of 7s, so what is left divides by 7 exactly when the number you started with does, and the final leftover is the true remainder.
Neither method is a digit rule, and that is not an oversight. The easy rules all come from a place value landing next to a multiple of the divisor. Ten is one more than 9, which is why adding digits works for 3 and 9. A hundred is one more than 99, which is nine 11s, which is why pairing digits works for 11.
10, 100 and 1000 leave awkward remainders when you share them into 7s: 3, 2 and 6 respectively. No tidy pattern appears in the digits, so no tidy rule exists.
Where it goes wrong
- Applying the rule for 3 by habit. Digit sums say nothing at all about 7.
- Stopping Chika's trick too early. Keep going until the result is inside the 7 times table you know.
- Reading the trick's answer as a remainder. Taking away 7s leaves the real remainder; the multiply-by-5 trick only ever answers yes or no.
- Treating the leftover as the answer. A leftover of 6 means the number is not divisible; it does not mean it divides by 6.
Questions people ask
Is there really no simple rule for 7?
Not one in the same class as 2, 5 or 10. Every published rule for 7 is an iterative procedure, which is why most curricula do not teach one at primary level.
Which method is best for a child?
Taking away 7s they already know. It uses times tables they have, it never goes negative, and it produces the remainder as a by-product.
Does Chika's trick work for other numbers?
The same idea works wherever a multiple of ten sits next to a multiple of the divisor. For 13 the multiplier is 4, because 40 is one more than 39.