To determine how many of the given numbers are divisible by 42, we need to check each number for divisibility by the prime factors of 42, which are 2, 3, and 7.
### Step-by-Step Solution:
1. **Identify the Numbers**: The numbers we need to check are:
- 2646
- 1008
- 1470
- 656
- 2478
- 1826
2. **Check for Divisibility by 2**: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).
- 2646 → last digit is 6 (divisible by 2)
- 1008 → last digit is 8 (divisible by 2)
- 1470 → last digit is 0 (divisible by 2)
- 656 → last digit is 6 (divisible by 2)
- 2478 → last digit is 8 (divisible by 2)
- 1826 → last digit is 6 (divisible by 2)
**Conclusion**: All numbers are divisible by 2.
3. **Check for Divisibility by 3**: A number is divisible by 3 if the sum of its digits is divisible by 3.
- 2646: 2 + 6 + 4 + 6 = 18 (divisible by 3)
- 1008: 1 + 0 + 0 + 8 = 9 (divisible by 3)
- 1470: 1 + 4 + 7 + 0 = 12 (divisible by 3)
- 656: 6 + 5 + 6 = 17 (not divisible by 3)
- 2478: 2 + 4 + 7 + 8 = 21 (divisible by 3)
- 1826: 1 + 8 + 2 + 6 = 17 (not divisible by 3)
**Conclusion**: The numbers divisible by 3 are 2646, 1008, 1470, and 2478.
4. **Check for Divisibility by 7**: A number is divisible by 7 if we can apply the rule: double the last digit, subtract it from the rest of the number, and check if the result is divisible by 7.
- 2646: 264 - (6 * 2) = 264 - 12 = 252 (252 ÷ 7 = 36, divisible)
- 1008: 100 - (8 * 2) = 100 - 16 = 84 (84 ÷ 7 = 12, divisible)
- 1470: 147 - (0 * 2) = 147 - 0 = 147 (147 ÷ 7 = 21, divisible)
- 2478: 247 - (8 * 2) = 247 - 16 = 231 (231 ÷ 7 = 33, divisible)
**Conclusion**: The numbers divisible by 7 are 2646, 1008, 1470, and 2478.
5. **Final Count**: The numbers that are divisible by 42 (which means they must be divisible by 2, 3, and 7) are:
- 2646
- 1008
- 1470
- 2478
Therefore, **4 numbers** are divisible by 42.
### Answer:
**4**
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