2014 AMC 12B Problem 10

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Concepts:place valuedivisibilityleast common multiple

Difficulty rating: 1680

10.

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 \ge 1 and a+b+c7.a+b+c \le 7. 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

Solution:

The distance driven is cbaabc=99(ca),cba - abc = 99(c-a), a multiple of 9.9. Driving a whole number of hours at 5555 mph makes it a multiple of 5555 too, hence a multiple of 495.495.

Since the odometer difference is at most a 33-digit number and a1,a \ge 1, the distance must be 495,495, so ca=5.c - a = 5.

With a1a \ge 1 and a+b+c7,a+b+c \le 7, the only choice is a=1,a=1, c=6,c=6, b=0.b=0. Then a2+b2+c2=1+0+36=37.a^2+b^2+c^2 = 1 + 0 + 36 = 37.

Thus, the correct answer is D.

Problem 10 in Other Years