Cindy wishes to arrange her coins into $X$ piles, each consisting of the same number of coins, $Y$. Each pile will have more than one coin and no pile will have all the coins. If there are 13 possible values for $Y$ given all of the restrictions, what is the smallest number of coins she could have?

Question

Gerald Dickerson

Business

15 August, 12:01

Cindy wishes to arrange her coins into $X$ piles, each consisting of the same number of coins, $Y$. Each pile will have more than one coin and no pile will have all the coins. If there are 13 possible values for $Y$ given all of the restrictions, what is the smallest number of coins she could have?

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Answers (1)

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Lorenzo Navarro · Answer 1

The smallest number of coins she could have is 4

Explanation:

With this Cindy respect all of the restrictions; Piles have the same number of coins, each pile will have more than one coin and no pile will have all the coins. That means X (piles) should be bigger than 1, and Y (coins in the piles) should also be bigger than 1. And there is no specification about the values of Y.

Therefore with a minimum of 4 coins, Cindy can do 2 piles of 2 coins each.