2014 AMC 10B Problem 14

Below is the professionally curated solution for Problem 14 of the 2014 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 10B solutions, or check the answer key.

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Concepts:digitsdivisibility

Difficulty rating: 1540

14.

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?

26 26

27 27

36 36

37 37

41 41

Solution:

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 .

Problem 14 in Other Years