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

2024 SSMO Speed Round Problems/Problem 2

Problem

Gracie's students play with some toys. When 4 or 5 students are present, the toys can be equally distributed to everyone. However, when there are only 3 students, there is one toy leftover after giving everyone the same number of toys. What is the least possible number of toys that Gracie could have?

Solution

Suppose that there $x$ toys. Since $x$ toys can be split equally when 4 of 5 students are present, $x$ must be a multiple of $\text{lcm}(4,5) = 20.$ So, $x$ can be $20,40,60,\dots.$ Now, since there is one toy left over when the toys are split when $3$ students are present, $x$ must be one greater than a multiple of $3.$ As $20 = 3\cdot6+2$ and $40 = 3\cdot13+1,$ we conclude that the answer is $\boxed{40}.$

~SMO_Team

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_SSMO_Speed_Round_Problems/Problem_2&oldid=260242"