Which of the following numbers is divisible by 99?
60687
To determine if a number is divisible by 99, we need to check if it is divisible by both 9 and 11, since $99 = 9 \times 11$.
Let's recall the divisibility rules for 9 and 11:
Let's apply these rules to each given option:
| Number | Sum of Digits | Divisible by 9? | Alternating Sum of Digits | Divisible by 11? | Divisible by 99? |
|---|---|---|---|---|---|
| 31548 | 21 | No | $8-4+5-1+3 = 11$ | Yes | No (Not divisible by 9) |
| 60687 | 27 | Yes | $7-8+6-0+6 = 11$ | Yes | Yes (Divisible by 9 and 11) |
| 44775 | 27 | Yes | $5-7+7-4+4 = 5$ | No | No (Not divisible by 11) |
| 84456 | 27 | Yes | $6-5+4-4+8 = 9$ | No | No (Not divisible by 11) |
Based on the checks, only the number 60687 is divisible by both 9 and 11, making it divisible by 99.
| Divisible By | Rule | Example |
|---|---|---|
| 2 | Ends in an even digit (0, 2, 4, 6, 8). | 48 (ends in 8) |
| 3 | Sum of digits is divisible by 3. | 123 ($1+2+3=6$, 6 is div by 3) |
| 4 | The number formed by the last two digits is divisible by 4. | 516 (16 is div by 4) |
| 5 | Ends in 0 or 5. | 75 (ends in 5) |
| 6 | Divisible by both 2 and 3. | 18 (even, $1+8=9$, 9 is div by 3) |
| 9 | Sum of digits is divisible by 9. | 729 ($7+2+9=18$, 18 is div by 9) |
| 10 | Ends in 0. | 150 (ends in 0) |
| 11 | Alternating sum of digits is divisible by 11. | 1331 ($1-3+3-1=0$, 0 is div by 11) |
When a number's divisor is a composite number (a number with more than two factors), like 99, we can often break down the divisibility check into its prime factors or relatively prime factors. Since $99 = 9 \times 11$, and 9 and 11 are relatively prime (they share no common factors other than 1), a number is divisible by 99 if and only if it is divisible by both 9 and 11.
It is important that the factors used are relatively prime. For instance, to check divisibility by 12, we check for divisibility by 3 and 4 (since 3 and 4 are relatively prime and $3 \times 4 = 12$), not 2 and 6 (since 2 and 6 are not relatively prime).
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