AMC12 2014 B
AMC12 2014 B · Q10
AMC12 2014 B · Q10. It mainly tests Rates (speed), Divisibility & factors.
Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \ge 1$ and $a + b + c \le 7$. At the end of the trip, the odometer showed $cba$ miles. What is $a^2 + b^2 + c^2$?
Danica 开着新车旅行了整数组小时,以平均 55 英里/小时的速度。旅行开始时,里程表显示 $abc$ 英里,其中 $abc$ 是三位数,$a \ge 1$ 且 $a + b + c \le 7$。旅行结束时,里程表显示 $cba$ 英里。求 $a^2 + b^2 + c^2$?
(A)
26
26
(B)
27
27
(C)
36
36
(D)
37
37
(E)
41
41
Answer
Correct choice: (D)
正确答案:(D)
Solution
Answer (D): Let $m$ be the total mileage of the trip. Then $m$ must be a multiple of $55$. Also, because $m=cba-abc=99(c-a)$, it is a multiple of $9$. Therefore $m$ is a multiple of $495$. Because $m$ is at most a 3-digit number and $a$ is not equal to $0$, $m=495$. Therefore $c-a=5$. Because $a+b+c\le 7$, the only possible $abc$ is $106$, so $a^2+b^2+c^2=1+0+36=37$.
答案(D):设 $m$ 为这次旅行的总里程,则 $m$ 必为 $55$ 的倍数。另外,因为 $m=cba-abc=99(c-a)$,所以它也是 $9$ 的倍数。因此 $m$ 是 $495$ 的倍数。由于 $m$ 至多是三位数且 $a\ne 0$,故 $m=495$。因此 $c-a=5$。又因为 $a+b+c\le 7$,唯一可能的 $abc$ 是 $106$,所以 $a^2+b^2+c^2=1+0+36=37$。
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