Subjects number theory

Digits A B 8B21Cc

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1. **Problem statement:** We are given a number 85AB1 which is a multiple of 99. We need to find the digits $A$ and $B$. 2. **Key fact:** A number is divisible by 99 if and only if it is divisible by both 9 and 11. 3. **Divisibility by 9 rule:** The sum of the digits must be divisible by 9. 4. **Divisibility by 11 rule:** The difference between the sum of digits in odd positions and the sum of digits in even positions must be divisible by 11. 5. Write the digits: $8, 5, A, B, 1$. 6. Sum of digits for divisibility by 9: $$8 + 5 + A + B + 1 = 14 + A + B$$ This sum must be divisible by 9. 7. For divisibility by 11, label positions from right to left: - Position 1 (rightmost): 1 (odd) - Position 2: B (even) - Position 3: A (odd) - Position 4: 5 (even) - Position 5: 8 (odd) Sum odd positions: $8 + A + 1 = 9 + A$ Sum even positions: $5 + B$ Difference: $$ (9 + A) - (5 + B) = 4 + A - B $$ This difference must be divisible by 11. 8. Let’s analyze divisibility by 9: Possible sums divisible by 9 near $14 + A + B$ are 18, 27, 36, ... Since $A$ and $B$ are digits (0 to 9), max sum is $14 + 9 + 9 = 32$, so possible sums are 18 or 27. 9. Case 1: $14 + A + B = 18 \\ A + B = 4$ 10. Case 2: $14 + A + B = 27 \\ A + B = 13$ 11. For divisibility by 11: $$4 + A - B = 0, \pm 11, \pm 22, ...$$ Since $A$ and $B$ are digits, difference is small, so consider: - $4 + A - B = 0 \\ \Rightarrow B = A + 4$ - $4 + A - B = 11 \\ \Rightarrow B = A - 7$ (not possible since $B$ must be ≥ 0) - $4 + A - B = -11 \\ \Rightarrow B = A + 15$ (not possible) 12. From $4 + A - B = 0$, we have $B = A + 4$. 13. Substitute into sums: - Case 1: $A + B = 4 \\ A + (A + 4) = 4 \\ 2A + 4 = 4 \\ 2A = 0 \\ A = 0, B = 4$ - Case 2: $A + B = 13 \\ A + (A + 4) = 13 \\ 2A + 4 = 13 \\ 2A = 9 \\ A = 4.5$ (not an integer digit, discard) 14. Therefore, the only solution is $A=0$ and $B=4$. 15. Check divisibility: Number: 85041 Sum digits: $8+5+0+4+1=18$ divisible by 9. Difference for 11: $(8+0+1)-(5+4) = 9 - 9 = 0$ divisible by 11. 16. Hence, $A=0$ and $B=4$ satisfy the conditions. **Final answer:** $A=0$, $B=4$