Difference between revisions of "2021 AIME I Problems/Problem 4"
(→Problem) |
Sugar rush (talk | contribs) |
||
| Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
| + | Suppose we have <math>1</math> coin in the first pile. Then <math>(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)</math> all work for a total of <math>31</math> piles. Suppose we have <math>2</math> coins in the first pile, then <math>(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)</math> all work, for a total of <math>29</math>. Continuing this pattern until <math>21</math> coins in the first pile, we have the sum <math>31+29+28+26+25+\ldots+4+2+1=(31+28+25+22+\ldots+1)+(29+26+23+\ldots+2)=176+155=\boxed{331}</math>. | ||
| + | |||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=3|num-a=5}} | {{AIME box|year=2021|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:54, 11 March 2021
Problem
Find the number of ways 
identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.
Solution
Suppose we have 
coin in the first pile. Then 
all work for a total of 
piles. Suppose we have 
coins in the first pile, then 
all work, for a total of 
. Continuing this pattern until 
coins in the first pile, we have the sum 
.
See also
| 2021 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
