Methods and worked examples: Number Play

2.1 Place value as algebra

Any number can be written in expanded place-value form. For example, 4075 = 4 × 1000 + 0 × 100 + 7 × 10 + 5. With letter-digits, a four-digit number dcba means 1000d + 100c + 10b + a, where a is the units digit, b the tens digit, c the hundreds digit and d the thousands digit.

This expansion explains every divisibility shortcut. The key fact is that each place value is a multiple of 10 except the units place: 10, 100, 1000 and so on. The second key fact is that each power of 10 is one more than a multiple of 9, and alternately one more or one less than a multiple of 11.

2.2 Rules for 2, 5 and 10

Since every digit except the units digit contributes a multiple of 10, the units digit alone decides divisibility by 10 and by 5. A number is divisible by 10 exactly when its units digit is 0. It is divisible by 5 exactly when its units digit is 0 or 5. It is divisible by 2 exactly when its units digit is 0, 2, 4, 6 or 8.

The same reasoning covers 4 and 8, using the last two and last three digits. Since 100 is a multiple of 4, the hundreds and higher places add multiples of 100, which are multiples of 4; only the last two digits matter. Since 1000 is a multiple of 8, only the last three digits matter for divisibility by 8. For example, 2856 ends in 56, which is divisible by 4, so 2856 is divisible by 4; its last three digits 856 are divisible by 8, so it is divisible by 8 too.

2.3 Divisibility by 9 and 3

Write a number as the sum of its digits plus a multiple of 9: N = (sum of digits) + 9 × (some whole number). This works because each place value 10, 100, 1000 is one more than 9, 99, 999, all multiples of 9. So N and the sum of its digits leave the same remainder when divided by 9. A number is divisible by 9 exactly when its digit total is a multiple of 9.

Example: 7309 has digit sum 7 + 3 + 0 + 9 = 19, then 1 + 9 = 10, then 1 + 0 = 1, so 7309 leaves remainder 1 on division by 9. The same reasoning with 10 = 9 + 1 and 9 divisible by 3 shows that a number is divisible by 3 exactly when its digit sum is divisible by 3.

2.4 Divisibility by 11

The place values alternate around multiples of 11: 1 = 0 + 1, 10 = 11 - 1, 100 = 99 + 1, 1000 = 1001 - 1, and so on. So the digits of a number, taken from the units place, alternate between adding and subtracting their place value.

The shortcut: add the digits in the odd positions counted from the units place, subtract the sum of the digits in the even positions, and check the result. If the difference is 0, 11 or a multiple of 11, the number is divisible by 11; otherwise the difference gives the remainder, adjusted to a positive value if needed. For 320185, the alternating sum is 5 - 8 + 1 - 0 + 2 - 3 = -3, which is 8 mod 11, so the number is 3 short of a multiple of 11 and leaves remainder 8.

2.5 Digital roots

The digital root of a number is the single digit left after adding its digits repeatedly. For 489710: 4 + 8 + 9 + 7 + 1 + 0 = 29, then 2 + 9 = 11, then 1 + 1 = 2, so the digital root is 2. The digital root is the remainder on division by 9, with one convention: when the remainder is 0, the digital root is written as 9. This is why every multiple of 9 has digital root 9.

Digital roots are useful as a quick check on arithmetic. If you add or multiply numbers, the digital root of the answer should match the digital root of the corresponding operation on the digital roots, with 9 standing for 0. Aryabhata II's Mahasiddhanta, from around 950 CE, describes this repeated digit-sum method and its use in checking calculations.

2.6 Cryptarithms

A cryptarithm is a puzzle in which letters stand for digits. Each letter is one digit, different letters are different digits, and the first digit of a multi-digit number is never 0. Addition and multiplication cryptarithms are solved by reasoning at the units place, carrying into the next column, and testing the small number of possible digits.

Worked example 3: A1 + 1B = B0. At the units place, 1 + B ends in 0, so B = 9 with a carry of 1. At the tens place, A + 1 + 1 = B = 9, so A = 7. Check: 71 + 19 = 90.

Worked example 4: AB + 37 = 6A. At the units place, B + 7 ends in A, with a carry of 1 to the tens place. At the tens place, A + 3 + 1 = 6, so A = 2. Then at the units place B + 7 ends in 2, so B = 5 (with the carry). Check: 25 + 37 = 62.

Worked example 5: PQ × 8 = RS. The product of a two-digit number with 8 is a two-digit number only when PQ is at most 12. The digits must differ, so PQ = 12 gives 12 × 8 = 96 with R = 9 and S = 6. PQ = 11 is excluded because the digits are equal, and 13 × 8 = 104 has three digits.