Difference between revisions of "2020 CIME I Problems/Problem 3"
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==Solution== | ==Solution== | ||
We see that the integers from 1 - 11 cannot be achieved. We similarly see that the integers from 16-23 cannot be reached. We then see that the integers from 31-35 cannot be reached. We then see that 46 and 47 cannot be reached. We then attempt to show that 61 can be reached. We see that 12x4 and 13 get the trick done. The answer is then 480. | We see that the integers from 1 - 11 cannot be achieved. We similarly see that the integers from 16-23 cannot be reached. We then see that the integers from 31-35 cannot be reached. We then see that 46 and 47 cannot be reached. We then attempt to show that 61 can be reached. We see that 12x4 and 13 get the trick done. The answer is then 480. | ||
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| + | ==See also== | ||
{{CIME box|year=2020|n=I|num-b=2|num-a=4}} | {{CIME box|year=2020|n=I|num-b=2|num-a=4}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
{{MAC Notice}} | {{MAC Notice}} | ||
Revision as of 16:54, 31 August 2020
Problem 3
In a math competition, all teams must consist of between 
and 
members, inclusive. Mr. Beluhov has 
students and he realizes that he cannot form teams so that each of his students is on exactly one team. Find the sum of all possible values of 
.
Solution
We see that the integers from 1 - 11 cannot be achieved. We similarly see that the integers from 16-23 cannot be reached. We then see that the integers from 31-35 cannot be reached. We then see that 46 and 47 cannot be reached. We then attempt to show that 61 can be reached. We see that 12x4 and 13 get the trick done. The answer is then 480.
See also
| 2020 CIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All CIME Problems and Solutions | ||
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 
