Q1.
The shortcut to find the divisibility by 3 is similar to the method for 9. A number is divisible by 3 if the sum of its digits is divisible by 3. Explore the remainders when powers of 10 are divided by 3. Explain why this method works.
Answer
Every power of 10 leaves a remainder of 1 when divided by 3.
| Power of 10 | Written as | Remainder ÷ 3 |
|---|---|---|
| 1 | 0 + 1 | 1 |
| 10 | 9 + 1 = 3 × 3 + 1 | 1 |
| 100 | 99 + 1 = 3 × 33 + 1 | 1 |
| 1000 | 999 + 1 = 3 × 333 + 1 | 1 |
| 10000 | 9999 + 1 = 3 × 3333 + 1 | 1 |
Why the method works: 9, 99, 999, 9999 are all multiples of 9 — and since 9 is itself a multiple of 3, they are all multiples of 3 as well. So splitting a number the same way as before,
… + 1000d + 100c + 10b + a
= (999d + 99c + 9b) + (d + c + b + a)
the first bracket is a multiple of 3, so the number and its digit sum leave the same remainder on division by 3.= (999d + 99c + 9b) + (d + c + b + a)
Examples: 15 → 1 + 5 = 6 ✓ and 15 = 3 × 5
87 → 8 + 7 = 15 ✓ and 87 = 3 × 29
429714 → 4+2+9+7+1+4 = 27 ✓ and 429714 = 3 × 143238
87 → 8 + 7 = 15 ✓ and 87 = 3 × 29
429714 → 4+2+9+7+1+4 = 27 ✓ and 429714 = 3 × 143238
Tip: this also explains why every multiple of 9 is a multiple of 3, but not the other way round. 15, 33 and 87 have digit sums 6, 6 and 15 — divisible by 3, not by 9.