Difference between revisions of "1988 AIME Problems/Problem 2"
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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. Since <math>1988</math> is even, we thus have <math>f_{1988}(11) = f_{4}(11) = \boxed{169}</math>. |
| − | <math>f_{ | ||
== See also == | == See also == | ||
| − | |||
| − | |||
{{AIME box|year=1988|num-b=1|num-a=3}} | {{AIME box|year=1988|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:16, 27 February 2018
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. Since 
is even, we thus have 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
