Difference between revisions of "1975 IMO Problems/Problem 4"
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<cmath>4444^{4444}<10000^{4444}=\left(10^4\right)^{4444}=10^{17776}</cmath> | <cmath>4444^{4444}<10000^{4444}=\left(10^4\right)^{4444}=10^{17776}</cmath> | ||
− | Therefore <math>4444^{4444}</math> has fewer than | + | Therefore <math>4444^{4444}</math> has fewer than 17776 digits. This shows that <math>A<9\cdot 17776=159984</math>. The sum of the digits of <math>A</math> is then maximized when <math>A=99999</math>, so <math>B\leq 45</math>. Note that out of all of the positive integers less than or equal to 45, the maximal sum of the digits is 12. |
It's not hard to prove that any base-10 number is equivalent to the sum of its digits modulo 9. Therefore <math>4444^{4444}\equiv A\equiv B(\bmod{9})</math>. This motivates us to compute <math>X</math>, where <math>1\leq X \leq 12</math>, such that <math>4444^{4444}\equiv X(\bmod{9})</math>. The easiest way to do this is by searching for a pattern. Note that | It's not hard to prove that any base-10 number is equivalent to the sum of its digits modulo 9. Therefore <math>4444^{4444}\equiv A\equiv B(\bmod{9})</math>. This motivates us to compute <math>X</math>, where <math>1\leq X \leq 12</math>, such that <math>4444^{4444}\equiv X(\bmod{9})</math>. The easiest way to do this is by searching for a pattern. Note that | ||
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Thus, <math>X=7</math>, which means that the sum of the digits of <math>B</math> is <math>\boxed{7}</math>. | Thus, <math>X=7</math>, which means that the sum of the digits of <math>B</math> is <math>\boxed{7}</math>. | ||
− | ~minor edits by KevinChen_Yay | + | |
+ | ~minor edits by [https://artofproblemsolving.com/wiki/index.php/User:Kevinchen_yay KevinChen_Yay] | ||
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{{alternate solutions}} | {{alternate solutions}} | ||
Latest revision as of 08:45, 24 April 2024
Problem
When is written in decimal notation, the sum of its digits is . Let be the sum of the digits of . Find the sum of the digits of . ( and are written in decimal notation.)
Solution
Note that
Therefore has fewer than 17776 digits. This shows that . The sum of the digits of is then maximized when , so . Note that out of all of the positive integers less than or equal to 45, the maximal sum of the digits is 12.
It's not hard to prove that any base-10 number is equivalent to the sum of its digits modulo 9. Therefore . This motivates us to compute , where , such that . The easiest way to do this is by searching for a pattern. Note that
and since ,
Thus, , which means that the sum of the digits of is .
~minor edits by KevinChen_Yay
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1975 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |