AMC10 2004 B

AMC10 2004 B · Q17

AMC10 2004 B · Q17. It mainly tests Linear equations, Remainders & modular arithmetic.

The two digits in Jack’s age are the same as the digits in Bill’s age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?

杰克的年龄的两数字与比尔的年龄的两数字相同，但顺序相反。五年后，杰克的年龄将是比尔当时年龄的两倍。他们当前年龄的差是多少？

(A) 9 9

(B) 18 18

(D) 36 36

(E) 45 45

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Let Jack’s age be $10x+y$ and Bill’s age be $10y+x$. In five years Jack will be twice as old as Bill. Therefore $10x+y+5=2(10y+x+5),$ so $8x=19y+5$. The expression $19y+5=16y+8+3(y-1)$ is a multiple of 8 if and only if $y-1$ is a multiple of 8. Since both $x$ and $y$ are 9 or less, the only solution is $y=1$ and $x=3$. Thus Jack is 31 and Bill is 13, so the difference between their ages is 18.

（B）设 Jack 的年龄为 $10x+y$，Bill 的年龄为 $10y+x$。五年后 Jack 的年龄将是 Bill 的两倍。因此 $10x+y+5=2(10y+x+5),$ 所以 $8x=19y+5$。表达式 $19y+5=16y+8+3(y-1)$ 是 8 的倍数当且仅当 $y-1$ 是 8 的倍数。由于 $x$ 和 $y$ 都不超过 9，唯一解为 $y=1$ 且 $x=3$。因此 Jack 31 岁，Bill 13 岁，他们年龄之差为 18。

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