2004 AIME II Problems/Problem 6
Problem
Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio 
what is the least possible total for the number of bananas?
Solution
Denote the number of bananas the first monkey took from the pile as 
, the second 
, and the third 
; the total is 
. Thus, the first monkey got 
, the second monkey got 
, and the third monkey got 
.
Taking into account the ratio aspect, say that the third monkey took 
bananas in total. Then,
Solve this to find that 
. All three fractions must be integral. Also note some other conditions we have picked up in the course of the problem, namely that is divisible by 
, is divisible by , and is divisible by 
(however, since the denominator contains a 
, the factors of 
cancel, and it only really needs to be divisible by ). Thus, the minimal value is when each fraction is equal to , and the solution is 
.
See also
| 2004 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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