+208 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 22 possible values for Y given all of the restrictions, what is the smallest number of coins she could have?
+622 The key to this problem lies in the fact that Cindy can have 22 different pile sizes (Y).
Here's why this information helps us find the smallest number of coins:
Divisibility: Since Cindy can have 22 different pile sizes, the total number of coins (let's call it T) must be divisible by 22 different numbers.
Smallest Divisors: The smaller the pile sizes (Y), the more divisors T will have. In other words, for T to have the smallest number of coins, we want the first 22 positive integers to divide T.
Smallest Multiple: The smallest number of coins (T) that satisfies all these conditions will be the least common multiple (LCM) of the first 22 positive integers.
Finding the LCM of all 22 numbers can be computationally expensive. However, we can make some observations:
All the first 11 positive integers divide any larger positive integer.
We only need to focus on the remaining prime numbers less than or equal to 22 (which are 2, 3, 5, 7, 11, 13, 17, and 19).
Checking Divisibility: We can check if T is divisible by each of these prime numbers. If it's not, we need to increase T until it becomes divisible by all of them.
Finding the Smallest T:
Start with a small number like T = 22 (which is divisible by 2 and 11). Check if it's divisible by 3, 5, 7, 13, 17, and 19. If not, increase T by 1 and repeat the checks.
The smallest T that satisfies all the divisibility conditions is T = 88.
Therefore, the smallest number of coins Cindy could have is 88.
