Question

Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had form a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts that each person has form an arithmetic progression. Find the number of peanuts Alex had initially.

答案

Solution:

After the eating, $444 - 5 - 9 - 25 = 405$ peanuts remain, and the three amounts form an arithmetic progression, so the middle amount, Betty's, is $\frac{405}{3} = 135.$ Hence Betty started with $135 + 9 = 144$ peanuts.

The starting amounts form a geometric progression, so they are $\frac{144}{r},$ $144,$ and $144r$ with $r \gt 1$ (Charlie had the most and Alex the least). Then $\frac{144}{r} + 144 + 144r = 444,$ which simplifies to $12r^2 - 25r + 12 = 0,$ with roots $r = \frac{4}{3}$ and $r = \frac{3}{4};$ since $r \gt 1,$ we take $r = \frac{4}{3}.$

So Alex initially had $144 \cdot \frac{3}{4} = 108$ peanuts. (Check: after eating, the amounts $103,$ $135,$ $167$ increase by $32$ each.)

2016 AIME II Problem 1

Solution:

Problem 1 in Other Years