Examine the following numbers for divisibility by 7:
5292758387

Answer

To determine if the number 5292758387 is divisible by 7, we can use the divisibility rule for 7. Here are the steps to follow:

### Step-by-Step Solution:

1. **Identify the last digit and the remaining number**:
- The last digit (unit digit) of the number 5292758387 is 7.
- The remaining number after removing the last digit is 529275838.

2. **Double the last digit**:
- Double of 7 is \( 7 \times 2 = 14 \).

3. **Subtract the doubled last digit from the remaining number**:
- Now, subtract 14 from 529275838:
\[ 529275838 - 14 = 529275824. \]

4. **Repeat the process**:
- The last digit of 529275824 is 4.
- The remaining number is 52927582.
- Double of 4 is \( 4 \times 2 = 8 \).
- Subtract 8 from 52927582:
\[ 52927582 - 8 = 52927574. \]

5. **Continue repeating**:
- The last digit of 52927574 is 4.
- The remaining number is 5292757.
- Double of 4 is \( 4 \times 2 = 8 \).
- Subtract 8 from 5292757:
\[ 5292757 - 8 = 5292749. \]

6. **Repeat again**:
- The last digit of 5292749 is 9.
- The remaining number is 529274.
- Double of 9 is \( 9 \times 2 = 18 \).
- Subtract 18 from 529274:
\[ 529274 - 18 = 529256. \]

7. **Continue the process**:
- The last digit of 529256 is 6.
- The remaining number is 52925.
- Double of 6 is \( 6 \times 2 = 12 \).
- Subtract 12 from 52925:
\[ 52925 - 12 = 52913. \]

8. **Repeat**:
- The last digit of 52913 is 3.
- The remaining number is 5291.
- Double of 3 is \( 3 \times 2 = 6 \).
- Subtract 6 from 5291:
\[ 5291 - 6 = 5285. \]

9. **Final steps**:
- The last digit of 5285 is 5.
- The remaining number is 528.
- Double of 5 is \( 5 \times 2 = 10 \).
- Subtract 10 from 528:
\[ 528 - 10 = 518. \]

10. **Final check**:
- The last digit of 518 is 8.
- The remaining number is 51.
- Double of 8 is \( 8 \times 2 = 16 \).
- Subtract 16 from 51:
\[ 51 - 16 = 35. \]

11. **Check if the final number is divisible by 7**:
- Now, we need to check if 35 is divisible by 7:
\[ 35 \div 7 = 5 \quad \text{(which is a whole number)}. \]

Since 35 is divisible by 7, we conclude that the original number 5292758387 is also divisible by 7.

### Final Conclusion:
The number 5292758387 is divisible by 7.