Difference between revisions of "1988 AIME Problems/Problem 2"
Mathcool2009 (talk | contribs) (→Solution) |
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== Solution == | == Solution == | ||
− | We see that <math> | + | We see that <math>f_{1}(11)=4</math> |
− | <math> | + | <math>f_2(11) = f_1(4)=16</math> |
− | <math> | + | <math>f_3(11) = f_1(16)=49</math> |
− | <math> | + | <math>f_4(11) = f_1(49)=169</math> |
− | <math> | + | <math>f_5(11) = f_1(169)=256</math> |
− | <math> | + | <math>f_6(11) = f_1(256)=169</math> |
Note that this revolves between the two numbers. | Note that this revolves between the two numbers. |
Revision as of 01:19, 6 June 2014
Problem
For any positive integer , let denote the square of the sum of the digits of . For , let . Find .
Solution
We see that
Note that this revolves between the two numbers.
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.