Question 1210674: 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?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(53979)   (Show Source):
You can put this solution on YOUR website! .
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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So, there are X piles and each pile has Y coins.
It means that the total number of coins is the product X*Y.
The restrictions are (a) Y > 1; (b) X > 1; (c) Y < X*Y.
The problem also states that the number X*Y has 13 possible divisors Y satisfying the imposed conditions.
It means that the total number of divisors of the number X*Y is 13 + 2 = 15
(divisors 1 and XY itself should be added to the 13 other divisors.)
The number 15 is a composite: it is the product of two prime numbers 15 = 3*5.
Due to the well-known formula for the number of divisors of a number, it means that
- one of the numbers X or Y has only two divisors greater than 1, which are some prime number of degree 1 and 2;
- the other of the two numbers X and Y is another prime number of degree 4.
The minimal such number, requested by the problem, is  = 9*16 = 144. ANSWER
Solved.
Answer by math_tutor2020(3842)  (Show Source):
You can put this solution on YOUR website!
x = number of piles
y = number of coins per pile
x,y are positive whole numbers.
xy = total number of coins
y > 1 since "each pile will have more than one coin"
x > 1 since the phrasing "each pile" implies there is more than one pile.
Or you can write it out like so,
y < total ... since "no pile will have all the coins"
y < xy
1 < x .... divide both sides by the variable y
x > 1
Since the total is xy, it's immediately evident that y is a factor of the total.
If there are 13 possible values for y, then that must mean the total xy has 13 different factors when ignoring 1 and xy itself.
We ignore 1 and xy because each pile has more than one coin, and we don't want all of the coins in one pile.
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Before continuing on, let's take a bit of a detour.
I'll explain how to find the number of factors for any integer.
The prime p has factors 1 and itself.
Powers of this prime, of the form p^k, have k+1 factors
Those factors being 1, p, p^2, ..., p^(k-1), p^k
For instance, 2^5 = 32 has 5+1 = 6 factors and they are 1,2,4,8,16,32
Consider p^k and q^m, where p & q are prime.
The product p^k*q^m has (k+1)(m+1) factors because we can make a table with k+1 rows and m+1 columns.
The k+1 rows represents the number of factors for p^k
The m+1 columns represents the number of factors for q^m
Check out this page for further reading
https://mathworld.wolfram.com/DivisorFunction.html
Refer specifically to equation (3) on that page.
That equation basically says "add 1 to each exponent, then multiply".
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Let's return to the current problem.
xy has 13 different factors when ignoring 1 and xy itself.
That gives 1+13+1 = 15 factors when including those values.
So we need to find a number that has 15 different factors.
Note that 15 = 3*5 can be written of the form (k+1)(m+1)
(k+1)(m+1) = 3*5
k+1 = 3 gives k = 2
m+1 = 5 gives m = 4
The order of k and m doesn't matter.
Next we select the smallest primes possible, so that we aim for the lowest coin count.
Select 2 and 3 to form 2^k*3^m = 2^2*3^4 = 4*81 = 324
But remember: the order of k and m doesn't matter, so we can easily swap the exponents.
Doing so will give
2^m*3^k = 2^4*3^2 = 16*9 = 144
The value 144 has 15 different factors and they are
1, 2, 3, 4, 6,
8, 9, 12, 16, 18,
24, 36, 48, 72, 144
That list was generated using GeoGebra's DivisorList function.
Many other online tools (such as WolframAlpha) can be used as an alternative.
So if you had 144 coins, then you can have...x = 2 piles and y = 72 coins per pile
x = 3 piles and y = 48 coins per pile
x = 4 piles and y = 36 coins per pile
x = 6 piles and y = 24 coins per pile
x = 8 piles and y = 18 coins per pile
x = 9 piles and y = 16 coins per pile
x = 12 piles and y = 12 coins per pile
x = 16 piles and y = 9 coins per pile
x = 18 piles and y = 8 coins per pile
x = 24 piles and y = 6 coins per pile
x = 36 piles and y = 4 coins per pile
x = 48 piles and y = 3 coins per pile
x = 72 piles and y = 2 coins per pile I used a custom Python script to generate that block of options above.
Note there are 13 lines in that block.
For any x,y pair of a given row, xy = 144.
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Answer: 144
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