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2014 AMC 10B Problem 14

Problem 14 of 25IntermediateNumber Theory

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 a≥1a\ge1 and a+b+c≤7.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?

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Solution

We know that the difference of the numbers cbacba and abcabc is equal to: 100c+10b+a−100a−10b−c100c + 10b+a - 100a - 10b-c =99(c−a)= 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 c−ac-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+c≤7a+ 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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Tagged: digits · divisibility

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