2014 AMC 10B 第 14 题

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14.

Danica 开着新车旅行了整数小时,平均速度为 5555 英里每小时。旅行开始时,里程表显示 abcabc 英里,其中 abcabc33 位数,a1a\ge1,且 a+b+c7a+b+c\le7。旅行结束时,里程表显示 cbacba 英里。求 a2+b2+c2a^2+b^2+c^2 的值。

Danica drove her new car on a trip for a whole number of hours, averaging 5555 miles per hour. At the beginning of the trip, abcabc miles was displayed on the odometer, where abcabc is a 33-digit number with a1a\ge1 and a+b+c7.a+b+c\le7. At the end of the trip, the odometer showed cbacba miles. What is a2+b2+c2?a^2+b^2+c^2?

2626

2727

3636

3737

4141

答案:D
知识点:数字整除性
难度评级:1540
解答:

里程表读数 cbacbaabcabc 之差为 ,而这个差必须是 5555 的倍数。因为 gcd(55,99)\gcd(55,99)1111,所以 cac-a55 的倍数,且 c>ac > a100c+10b+a100a10bc100c + 10b+a - 100a - 10b-c =99(ca)= 99(c-a)

a+b+c7a+ b+c \leq 7 的限制下,唯一可能是 a=1,b=0,c=6a = 1, b = 0, c = 6,其他组合都会有 a+b+c>7a+b+c > 7。因此 a2+b2+c2=37a^2+b^2+c^2 = 37

所以正确答案是 D

We know that the difference of the numbers cbacba and abcabc is equal to: 100c+10b+a100a10bc100c + 10b+a - 100a - 10b-c =99(ca)= 99(c-a) We know that this number also must be a multiple of 55.55. As gcd(55,99)\gcd(55,99) is 11,11, we know that cac-a is a multiple of 5,5, and c>a.c > a.

This makes a=1,b=0,c=6a = 1, b = 0, c = 6 the only possible value with a+b+c7a+ b+c \leq 7 as every other combination has a+b+c>7.a+b+c > 7. As such, a2+b2+c2=37.a^2+b^2+c^2 = 37.

Thus, the correct answer is D .

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