2010 AMC 10A Problem 25

Attempt Problem 25 of the 2010 AMC 10A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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: 5555−72=66−22=22−12=11−12=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
Answer: B
Concepts:perfect squarerecursionwork backwards
Small hint:
Build the smallest possible sequence backwards from 00
Big hint:
At each reverse step, add the smallest square that the greedy rule would subtract
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. Continuing until there are eight terms in the forward sequence gives 7223.7223.
72237223−842=167167−122=2323−42=77−22=33−12=22−12=11−12=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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