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2015 AMC 12B Problem 21

Problem 21 of 25HarderAlgebra

Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 55 steps left). Suppose that Dash takes 1919 fewer jumps than Cozy to reach the top of the staircase. Let ss denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of s?s?

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Solution

A staircase of tt steps takes Cozy t2\left\lceil \tfrac{t}{2} \right\rceil jumps and Dash t5\left\lceil \tfrac{t}{5} \right\rceil jumps, and we need the difference to equal 19.19. Suppose Dash makes d+1d+1 jumps. Then tt is one of 5d+1,5d+1, 5d+2,5d+2, 5d+3,5d+3, 5d+4,5d+4, 5d+5.5d+5. Cozy makes d+20d+20 jumps, so tt is either 2d+392d+39 or 2d+40.2d+40. Equating these two lists gives an integer dd only in the three cases 5d+3=2d+39,5d+1=2d+40,5d+4=2d+40. \begin{gathered} 5d+3=2d+39,\\ 5d+1=2d+40,\\ 5d+4=2d+40. \end{gathered} These yield respectively t=63,66,64.t=63,66,64. Thus the valid values are t=63,t=63, 64,64, and 66,66, so s=63+64+66=193.s = 63 + 64 + 66 = 193. Its digit sum is 1+9+3=13.1 + 9 + 3 = 13. Thus, the correct answer is D.

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Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.