Why the digit-sum test works
Tonight's idea
A number is divisible by 9 exactly when its digits add to a multiple of 9. Why should digits know anything about dividing?
Because 10 is one more than 9: \(100 = 99 + 1\), \(1000 = 999 + 1\), and every power of ten is a multiple of 9 plus 1.
Take \(752 = 7 \times 100 + 5 \times 10 + 2\) and swap in that fact:
\[752 = 7(99 + 1) + 5(9 + 1) + 2 = (7 \times 99 + 5 \times 9) + (7 + 5 + 2).\]
The first bracket is a multiple of 9 whatever the digits are. So 752 and its digit sum 14 differ by a multiple of 9: they leave the same remainder on dividing by 9, and one is a multiple of 9 exactly when the other is. Since 9, 99, 999, … are also multiples of 3, the same argument gives the test for 3.
For 11, 10 is one less than 11: \(10 = 11 - 1\), \(100 = 9 \times 11 + 1\), \(1000 = 91 \times 11 - 1\). Powers of ten alternate between one below and one above a multiple of 11, so for digits \(a, b, c, d\):
\(1000a + 100b + 10c + d = (\text{multiple of } 11) - a + b - c + d.\)
Alternately add and subtract the digits from the right. If the result is a multiple of 11 (0 counts), so is the number.
Watch one: 918,082 under both tests
Digit sum: \(9 + 1 + 8 + 0 + 8 + 2 = 28\), not a multiple of 3 or 9, so 918,082 is divisible by neither. Better: its remainder on dividing by 9 equals the remainder of 28, which is 1.
For 11, start at the right and alternate: \(2 - 8 + 0 - 8 + 1 - 9 = -22\), a multiple of 11. So 918,082 is divisible by 11. Check: \(11 \times 83{,}462 = 918{,}082\).
Remainders are the subject of clock arithmetic: a number and its digit sum sit at the same place on a 9-hour clock.
Your turn
- Without dividing, say whether each of 7,254, 48,627 and 52,314 is divisible by 3, and whether it is divisible by 9.
- (a) Find the digit \(\square\) that makes \(5\square 31\) divisible by 9. There are two answers. (b) Find every digit \(\square\) that makes \(7\square 2\) divisible by 3.
- Apply the alternating test to 2,728, to 1,353 and to 10,000,001. Then explain, using \(10 = 11 - 1\), why 10,000,001 was certain to pass.
- Casting out nines. A student claims \(347 \times 268 = 92{,}986\). Replace 347 and 268 by the remainders of their digit sums on dividing by 9, multiply those, and reduce again. Do the same to the claimed answer and show the claim is wrong without multiplying. Why does this work? Name one mistake it can never catch.
Stretch: blocks of three
\(1001 = 7 \times 11 \times 13\), and \(1000 = 1001 - 1\). So the alternating trick, applied to blocks of three digits instead of single digits, tests for 7, 11 and 13 at once. Is 1,234,562 divisible by 7? By 13? Explain why the block method is allowed.
Answers
- 7,254: digit sum 18, divisible by 3 and by 9. 48,627: digit sum 27, divisible by both. 52,314: digit sum 15, divisible by 3 but not by 9.
- (a) \(5 + \square + 3 + 1 = 9 + \square\) must be a multiple of 9, so \(\square = 0\) or \(9\): 5,031 and 5,931. (b) \(9 + \square\) must be a multiple of 3, so \(\square = 0, 3, 6\) or \(9\).
- 2,728: \(8 - 2 + 7 - 2 = 11\), divisible. 1,353: \(3 - 5 + 3 - 1 = 0\), divisible (\(11 \times 123\)). 10,000,001: \(1 - 0 + 0 - 0 + 0 - 0 + 0 - 1 = 0\), divisible. Why: \(10^7\) is an odd power of \(11 - 1\), so it is a multiple of 11 minus 1, and adding 1 gives a multiple of 11. In fact \(10{,}000{,}001 = 11 \times 909{,}091\).
- 347 has digit sum 14, remainder 5. 268 has digit sum 16, remainder 7. \(5 \times 7 = 35\), remainder 8. So the true product leaves remainder 8 on dividing by 9. The claim 92,986 has digit sum 34, remainder 7. Wrong. (The product is 92,996.) It works because a number and its digit sum have the same remainder on dividing by 9, and remainders multiply. It can never catch two digits swapped, since the digit sum does not change.
Stretch. Blocks from the right: \(562 - 234 + 1 = 329 = 7 \times 47\), so 1,234,562 is divisible by 7. \(329 = 13 \times 25 + 4\), so it is not divisible by 13. The method is allowed because \(1000 = 1001 - 1\) and \(1000^2 = 1001 \times 999 + 1\): powers of 1000 alternate between one less and one more than a multiple of 1001, exactly as powers of 10 do for 11.