AMC12 2012 B

AMC12 2012 B · Q11

AMC12 2012 B · Q11. It mainly tests Linear equations, Base representation.

In the equation below, $A$ and $B$ are consecutive positive integers, and $A$, $B$, and $A+B$ represent number bases: \[132_A+43_B=69_{A+B}.\] What is $A+B$?

在下面的方程中，$A$ 和 $B$ 是连续的正整数，且 $A$、$B$ 和 $A+B$ 表示进制：\[132_A+43_B=69_{A+B}.\] $A+B$ 是多少？

(A) 9 9

(B) 11 11

(D) 15 15

(E) 17 17

Answer

Correct choice: (C)

正确答案：（C）

Solution

Change the equation to base 10: \[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\] \[A^2 - 3A - 2B - 4=0\] Either $B = A + 1$ or $B = A - 1$, so either $A^2 - 5A - 6, B = A + 1$ or $A^2 - 5A - 2, B = A - 1$. The second case has no integer roots, and the first can be re-expressed as $(A-6)(A+1) = 0, B = A + 1$. Since A must be positive, $A = 6, B = 7$ and $A+B = \boxed{\textbf{(C)}\ 13}$

将等式转换为十进制：\[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\] \[A^2 - 3A - 2B - 4=0\] 由于 $B=A+1$ 或 $B=A-1$，所以要么 $A^2 - 5A - 6, B = A + 1$，要么 $A^2 - 5A - 2, B = A - 1$。第二种情况没有整数根；第一种情况可改写为 $(A-6)(A+1) = 0, B = A + 1$。由于 $A$ 必须为正，得 $A = 6, B = 7$，因此 $A+B = \boxed{\textbf{(C)}\ 13}$

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