2010 AMC 10A Problems/Problem 25
Problem
Jim starts with a positive integer 
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 
, then his sequence contains 
numbers:
![\[\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}\]](http://latex.artofproblemsolving.com/2/e/d/2eda46643f3b6313f65bb202d501f3391552655b.png)
Let 
be the smallest number for which Jim’s sequence has 
numbers. What is the units digit of ?
Solution
We can find the answer by working backwards. We begin with 
on the bottom row, then the 
goes to the right of the equal's sign in the row above. We find the smallest value 
for which 
and 
, which is 
.
We repeat the same procedure except with for the next row and 
for the row after that. However, at the fourth row, we see that solving 
yields 
, in which case it would be incorrect since 
is not the greatest perfect square less than or equal to . So we make it a 
and solve 
. We continue on using this same method where we increase the perfect square until can be made bigger than it. When we repeat this until we have rows, we get:
![\[\begin{array}{ccccc}{}&{}&{}&{}&7223\\ 7223&-&84^{2}&=&167\\ 167&-&12^{2}&=&23\\ 23&-&4^{2}&=&7\\ 7&-&2^{2}&=&3\\ 3&-&1^{2}&=&2\\ 2&-&1^{2}&=&1\\ 1&-&1^{2}&=&0\\ \end{array}\]](http://latex.artofproblemsolving.com/8/a/c/8acd99dc227327e964c7e0b92b1319c372a9e983.png)
Hence the solution is the last digit of 
, which is 
.
See also
| 2010 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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