To find the greatest value of such that the 7-digit number 485A64B is divisible by both 8 and 9, we follow these steps:
Step 1: Divisibility by 8
A number is divisible by 8 if its last three digits are divisible by 8. Here, the last three digits are 64B. We need 64B to be divisible by 8. Let's test values of .
Checking divisibility for each :
- B = 0: 640/8 = 80 (divisible)
- B = 1: 641/8 = 80.125 (not divisible)
- B = 2: 642/8 = 80.25 (not divisible)
- B = 3: 643/8 = 80.375 (not divisible)
- B = 4: 644/8 = 80.5 (not divisible)
- B = 5: 645/8 = 80.625 (not divisible)
- B = 6: 646/8 = 80.75 (not divisible)
- B = 7: 647/8 = 80.875 (not divisible)
- B = 8: 648/8 = 81 (divisible)
- B = 9: 649/8 = 81.125 (not divisible)
The possible values of that make 64B divisible by 8 are 0 and 8.
Step 2: Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. The sum of the digits in 485A64B is .
Check divisibility for each that works from Step 1:
- B = 0: 27 + A + 0 must be divisible by 9.
- B = 8: 27 + A + 8 = 35 + A must be divisible by 9.
For B = 0: must be 36, 45, etc., so possible = 9. Then,
For B = 8: must be 36, 45, etc., so possible = 1. Then,
Thus, the greatest value of that satisfies both conditions is .
The correct answer is .