Problem:
Initially Alex, Betty, and Charlie had a total of 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 of his peanuts, Betty eats of her peanuts, and Charlie eats 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.
Solution:
After the three people eat the peanuts, peanuts remain. Hence after eating the peanuts, for some positive integers and , Alex, Betty, and Charlie have , and peanuts, respectively. Then , and . Thus Betty originally had peanuts. Because the initial numbers of peanuts were in geometric progression, for some , Alex originally had peanuts and Charlie originally had peanuts. Because there was originally a total of peanuts, it follows that . The only solution greater than is . Alex initially had peanuts.
The problems on this page are the property of the MAA's American Mathematics Competitions