Difference between revisions of "1988 AIME Problems/Problem 2"
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== Problem == | == Problem == | ||
+ | For any positive integer <math>k</math>, let <math>f_1(k)</math> denote the square of the sum of the digits of <math>k</math>. For <math>n \ge 2</math>, let <math>f_n(k) = f_1(f_{n - 1}(k))</math>. Find <math>f_{1988}(11)</math>. | ||
== Solution == | == Solution == | ||
− | We see that <math> | + | We see that <math>f_{1}(11)=4</math> |
− | <math> | + | |
− | <math> | + | <math>f_2(11) = f_1(4)=16</math> |
− | <math> | + | |
− | <math> | + | <math>f_3(11) = f_1(16)=49</math> |
− | <math> | + | |
− | Note that this revolves between the two numbers. | + | <math>f_4(11) = f_1(49)=169</math> |
− | <math>f_{ | + | |
+ | <math>f_5(11) = f_1(169)=256</math> | ||
+ | |||
+ | <math>f_6(11) = f_1(256)=169</math> | ||
+ | |||
+ | 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>. | ||
== 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 |
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