2010 AMC 10A Problem 25

Below is the professionally curated solution for Problem 25 of the 2010 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10A solutions, or check the answer key.

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Concepts:perfect squarerecursionwork backwards

Difficulty rating: 2440

25.

Jim starts with a positive integer nn and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with n=55,n = 55, then his sequence contains 55 numbers: 555572=6622=2212=1112=0\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array} Let NN be the smallest number for which Jim’s sequence has 88 numbers. What is the units digit of N?N?

11

33

55

77

99

Solution:

We can work backwards starting with 0.0. From this, we can add on 121^2 to get 1.1.

We can again add on 121^2 to get 2.2. Again, adding on 121^2 gives us 3.3.

If we add on 121^2 now, we get 4,4, but then 121^2 is not the greatest square less than or equal to 4.4.

Then adding on 222^2 gives us 7.7. We repeat this process for 88 steps to get 7223.7223.

72237223842=167167122=232342=7722=3312=2212=1112=0\begin{array}{ccccc} {}&{}&{}&{}&7223\\ 7223&-&84^2&=&167\\ 167&-&12^2&=&23\\ 23&-&4^2&=&7\\ 7&-&2^2&=&3\\ 3&-&1^2&=&2\\2&-&1^2&=&1\\1&-&1^2&=&0\end{array}

Thus, B is the correct answer.

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