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 "2024 SSMO Team Round Problems/Problem 2"

(Created page with "==Problem== Find the sum of the three smallest positive integers <math>n</math> where the last two digits of <math>n^4</math> are <math>01</math>. ==Solution==")
 
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
From <math>n^4\equiv1\pmod{100},</math> we have <math>n^4\equiv1\pmod{4}</math> and <math>n^4\equiv1\pmod{25}.</math> From the first congruence, we have <math>n\equiv1\pmod{2}.</math> Now, let <math>n\equiv5a+b\pmod{25}</math> for integers <math>0\le a,b< 5.</math> Then, <cmath>(5a+b)^4 = 5^4a^4+4\cdot5^3a^3b+6\cdot5^2a^2b^2+4\cdot5ab^3+b^4 \equiv 20ab^3+b^4\pmod{25}</cmath> so <cmath>20ab^3+b^4 \equiv 1 \pmod{25}.</cmath> Now, we will proceed using casework. Clearly, <math>b</math> is not a multiple of 5. Now, note that
 +
\begin{align*}
 +
b\equiv1\pmod{5}&\implies a\equiv0\pmod{5},\\
 +
b\equiv2\pmod{5}&\implies a\equiv1\pmod{5},\\
 +
b\equiv3\pmod{5}&\implies a\equiv3\pmod{5},\text{ and }\\
 +
b\equiv4\pmod{5}&\implies a\equiv4\pmod{5}.
 +
\end{align*}
 +
So, we have <math>5a+b \in \{1,7,18,24\}.</math> Combining this with <math>n\equiv1\pmod{2},</math> we find the following solutions as residues mod 50: <math>1,7,43,</math> and <math>49.</math> So, the answer is <math>1+7+43 = \boxed{51}.</math>
 +
 +
~SMO_Team

Latest revision as of 14:36, 10 September 2025

Problem

Find the sum of the three smallest positive integers $n$ where the last two digits of $n^4$ are $01$.

Solution

From $n^4\equiv1\pmod{100},$ we have $n^4\equiv1\pmod{4}$ and $n^4\equiv1\pmod{25}.$ From the first congruence, we have $n\equiv1\pmod{2}.$ Now, let $n\equiv5a+b\pmod{25}$ for integers $0\le a,b< 5.$ Then, \[(5a+b)^4 = 5^4a^4+4\cdot5^3a^3b+6\cdot5^2a^2b^2+4\cdot5ab^3+b^4 \equiv 20ab^3+b^4\pmod{25}\] so \[20ab^3+b^4 \equiv 1 \pmod{25}.\] Now, we will proceed using casework. Clearly, $b$ is not a multiple of 5. Now, note that \begin{align*} b\equiv1\pmod{5}&\implies a\equiv0\pmod{5},\\ b\equiv2\pmod{5}&\implies a\equiv1\pmod{5},\\ b\equiv3\pmod{5}&\implies a\equiv3\pmod{5},\text{ and }\\ b\equiv4\pmod{5}&\implies a\equiv4\pmod{5}. \end{align*} So, we have $5a+b \in \{1,7,18,24\}.$ Combining this with $n\equiv1\pmod{2},$ we find the following solutions as residues mod 50: $1,7,43,$ and $49.$ So, the answer is $1+7+43 = \boxed{51}.$

~SMO_Team

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_SSMO_Team_Round_Problems/Problem_2&oldid=260264"