Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1975 IMO Problems/Problem 4"

m (→‎Solution)
m (→‎Solution)
Line 7: Line 7:
 
<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 17776 digits. This shows that <math>A<9\cdot 17775=159975</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.
+
Therefore <math>4444^{4444}</math> has fewer than 17777 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
Line 19: Line 19:
 
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
 
{{alternate solutions}}
 
{{alternate solutions}}
  

Revision as of 18:15, 26 May 2023

Problem

When $4444^{4444}$ is written in decimal notation, the sum of its digits is $A$. Let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$. ($A$ and $B$ are written in decimal notation.)

Solution

Note that \[4444^{4444}<10000^{4444}=\left(10^4\right)^{4444}=10^{17776}\]

Therefore $4444^{4444}$ has fewer than 17777 digits. This shows that $A<9\cdot 17776=159984$. The sum of the digits of $A$ is then maximized when $A=99999$, so $B\leq 45$. 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 $4444^{4444}\equiv A\equiv B(\bmod{9})$. This motivates us to compute $X$, where $1\leq X \leq 12$, such that $4444^{4444}\equiv X(\bmod{9})$. The easiest way to do this is by searching for a pattern. Note that

\[4444^1\equiv 7(\bmod 9)\\4444^2\equiv 4(\bmod 9)\\4444^3\equiv 1(\bmod 9)\]

and since $4444=3\times 1481+1$,

\[4444^{4444}\equiv 4444^{3\times1481+1}\equiv \left(4444^3\right)^{1481}\times 4444\equiv 1\times 4444\equiv 7(\bmod{9})\]

Thus, $X=7$, which means that the sum of the digits of $B$ is $\boxed{7}$.

~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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1975_IMO_Problems/Problem_4&oldid=193546"