Difference between revisions of "2024 SSMO Accuracy Round Problems/Problem 3"
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==Solution== | ==Solution== | ||
| + | Note that <math>1!,2!,\dots,10!</math> have residues <math>1,2,1,4,0,0,\dots</math> when taken modulo 5. We have 2 cases for residues modulo 5: either one is 0, one is 1, and one is 4, or they are all 0. For the first case, we have <math>2</math> ways to choose a <math>1</math>, 1 way to choose a <math>4,</math> <math>6</math> ways to choose a <math>0,</math> and <math>6</math> ways to permute them, giving <math>2\cdot6\cdot6 = 72</math> solutions. For the other case, we have <math>6\cdot5\cdot4 = 120</math> ways to choose the 3 multiples of 5. Thus, the total number of solutions is <math>120+72 = 192,</math> meaning the answer is <cmath>\frac{192}{10 \cdot 9 \cdot 8} = \frac{4}{15}\implies 4+15 = \boxed{19}.</cmath> | ||
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| + | ~SMO_Team | ||
Latest revision as of 14:31, 10 September 2025
Problem
Three distinct random integers 
, 
, and 
are selected so that 
. Let the probability that 
be 
, where 
. Find 
.
Solution
Note that 
have residues 
when taken modulo 5. We have 2 cases for residues modulo 5: either one is 0, one is 1, and one is 4, or they are all 0. For the first case, we have 
ways to choose a 
, 1 way to choose a 
ways to choose a 
and ways to permute them, giving 
solutions. For the other case, we have 
ways to choose the 3 multiples of 5. Thus, the total number of solutions is 
meaning the answer is ![\[\frac{192}{10 \cdot 9 \cdot 8} = \frac{4}{15}\implies 4+15 = \boxed{19}.\]](http://latex.artofproblemsolving.com/b/1/3/b13d47390033781de94cb973f5aef9c29c32f7e5.png)
~SMO_Team
