How to tell if a number is divisible by 13
This is long division said out loud: 13 × 90 is 1170, take it away, look at what is left. There is nothing to memorise, and it works on any number, however long.
Take away multiples of 13 you already know until what is left is small enough to recognise from the 13 times table.
How to check it
- Find a multiple of 13 that fits: 13 × 100 = 1300, 13 × 10 = 130.
- Take it away and look at what is left.
- Repeat until the leftover is inside the 13 times table.
- Landing on 0 also means the number divides by 13.
Worked examples
Is 1183 divisible by 13?
1183 − 1170 (13 × 90) = 13 13 = 13 × 1
Nothing left over, so yes. 1183 = 13 × 91.
Is 1147 divisible by 13?
1147 − 1040 (13 × 80) = 107 107 − 104 (13 × 8) = 3
3 left over, so no. 1147 = 13 × 88 remainder 3.
Chop, multiply by 4, add
Chop off the last digit, multiply it by 4, and add it to what is left. Repeat until the number is small enough to judge.
1183 → 118 + 3 × 4 = 130 → 13 + 0 × 4 = 13, which is 13 × 1. So 1183 divides by 13.
It works because 40 is one more than 39, and 39 is three 13s. Multiplying the last digit by 4 and adding it back does not change whether the number divides by 13. It does change what is left over, so this method answers yes or no and never gives the remainder.
Why these methods work
Taking away 13s cannot change the answer. Every chunk that comes off is a whole number of 13s, so what is left divides by 13 exactly when the number you started with does, and the final leftover is the true remainder.
Neither method is a digit rule, because a digit rule needs a place value landing next to a multiple of the divisor. For 13, ten leaves 10, a hundred leaves 9 and a thousand leaves 12, so nothing lines up.
Where mistakes happen
- Not being familiar with the multiples of 13. Build the ones you need: e.g. 13 × 10 = 130, take away one 13 for 13 × 9 = 117, then ten of those for 13 × 90 = 1170.
Questions people ask
Which method should I use?
Taking away 13s. It gives you the remainder as well as the yes or no, and it never needs a multiple you cannot build.
Where is this rule useful?
Breaking a number into prime factors, and simplifying fractions.