Difference between revisions of "2020 CIME I Problems/Problem 3"
Magnetoninja (talk | contribs) (→Solution 2 (Methodic Process)) |
|||
| (3 intermediate revisions by 2 users not shown) | |||
| Line 4: | Line 4: | ||
==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. | ||
| − | |||
| + | == Solution 2 (Methodic Process) == | ||
| + | |||
| + | Let <math>t</math> be the total number of students in his class. Then, that means that <math>t\pmod{12}>3\lfloor{\frac{t}{12}}\rfloor</math> | ||
| + | This translates to when there are <math>12</math> students on a team, there are more remaining students than slots on those teams, where each team has a maximum of three slots for <math>\lfloor{\frac{t}{12}}\rfloor</math> teams. Those three slots are the <math>13</math>th <math>14</math>th and <math>15</math>th members of the team. We see that <math>3\lfloor{\frac{t}{12}}\rfloor</math> has to be less than <math>12</math>, so <math>\lfloor{\frac{t}{12}}\rfloor</math> is between <math>0</math> and <math>3</math> inclusive. Plugging in <math>x= {0,1,2,3,}</math> give the numbers <math>1-11</math>, <math>16-23</math>, <math>31-35</math>, and <math>46-47</math> respectively. Adding these up, we get <math>\boxed{480}</math> | ||
| + | |||
| + | -Magnetoninja | ||
| + | |||
| + | ==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}} | ||
Latest revision as of 20:48, 17 May 2023
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.
Solution 2 (Methodic Process)
Let 
be the total number of students in his class. Then, that means that 
This translates to when there are students on a team, there are more remaining students than slots on those teams, where each team has a maximum of three slots for 
teams. Those three slots are the 
th 
th and th members of the team. We see that 
has to be less than , so is between 
and 
inclusive. Plugging in 
give the numbers 
, 
, 
, and 
respectively. Adding these up, we get 
-Magnetoninja
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. 
