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Find the smallest possible values of A and B for the number 645A2879B8 to be divisible by both 8 and 9.
Published on November 11, 2024
Pre-Algebra
To find the smallest possible values for A and B in a 10-digit number divisible by both 8 and 9, the last three digits (9B8) must be divisible by 8, and the sum of all digits must be divisible by 9. This leads to B=2 and A=3 as the smallest possible values.
Question
Answer
Here’s how to solve this problem:
Divisibility by 8:
A number is divisible by 8 if its last three digits form a number divisible by 8. In our case, the last three digits are 9B8. We need to find a value for B such that 9B8 is divisible by 8.
- If B = 2, then 928 is divisible by 8 (928/8 = 116).
- If B = 6, then 968 is divisible by 8 (968/8 = 121).
Since we are looking for the smallest value of B, we choose B = 2.
Divisibility by 9:
A number is divisible by 9 if the sum of its digits is divisible by 9. The digits of our number are 6, 4, 5, A, 2, 8, 7, 9, B, 8. We already determined B = 2. So the sum of the digits is:
6 + 4 + 5 + A + 2 + 8 + 7 + 9 + 2 + 8 = 51 + A
We need 51 + A to be divisible by 9. The smallest possible value for A that makes this true is A = 3, since 51 + 3 = 54, and 54 is divisible by 9 (54/9 = 6).
Conclusion:
The smallest possible values are A = 3 and B = 2. Therefore, the correct answer is option 3.
Final Answer: The final answer is