Difference between revisions of "2024 SSMO Speed Round Problems/Problem 2"
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| + | Suppose that there <math>x</math> toys. Since <math>x</math> toys can be split equally when 4 of 5 students are present, <math>x</math> must be a multiple of <math>\text{lcm}(4,5) = 20.</math> So, <math>x</math> can be <math>20,40,60,\dots.</math> Now, since there is one toy left over when the toys are split when <math>3</math> students are present, <math>x</math> must be one greater than a multiple of <math>3.</math> As <math>20 = 3\cdot6+2</math> and <math>40 = 3\cdot13+1,</math> we conclude that the answer is <math>\boxed{40}.</math> | ||
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| + | ~SMO_Team | ||
Latest revision as of 14:25, 10 September 2025
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 
toys. Since toys can be split equally when 4 of 5 students are present, must be a multiple of 
So, can be 
Now, since there is one toy left over when the toys are split when 
students are present, must be one greater than a multiple of 
As 
and 
we conclude that the answer is 
~SMO_Team
