A standard deck of playing cards with $26$ red cards and $26$ black cards is split into two piles, each having at least one card. In pile $A$ there are six times as many black cards as red cards. In pile $B,$ the number of red cards is a multiple of the number of black cards. How many red cards are in pile $B

We have 26 red cards and 26 black cards.

we have two piles, each with at least one card.

Lets call Ar at the number of red cards in pile A, Ab as the numer of black cards in pile A, Br as the number of red cards in the pile B and Bb as the number of black cards in pile B.

The equations that we have are:

Ar + Br = 26

Ab + Bb = 26

Ab = 6Ar

Br = k*Bb

where k is a constant, that gives us a multiple of the number of black cards in B.

Now, lets solve this:

first, in the two first equations we can isolate two variables and get:

Ar = 26 - Br

Ab = 26 - Bb

now, replace this in the other two equations:

(26 - Bb) = 6 * (26 - Br)

Br = k*Bb

now, we can replace the second equation into the first:

(26 - Bb) = 6 * (26 - Br) = 6 * (26 - k*Bb)

now we solve it for Bb

26 - Bb = 6*26 - 6*k*Bb

Bb * (6*k - 1) = 6*26 - 26 = 5*26

Bb * (6k - 1) = 130

Bb = 130 / (6*k - 1)

Now, remember that Bb can only be a whole number, So 6k - 1 must divide 130:

if K = 1, we have:

Bb = 130 / (6 - 1) = 130/5 = 26

so the black cards are located in the B pile, this has no sense, so k can not be equal to 1.

Let's try with k = 2.

Bb = 130 / (12 - 1) = 130/11 = 11.8, this is not a whole number:

k = 3

Bb = 130 / (6*3 - 1) = 130 / (17) = 7.6, this is not a whole number.

You can keep triyng, you will find that when k = 11

Bb = 130 / (6*11 - 1) = 2

So we have 2 black cards in the pile B, then we have 11*2 = 22 red cards in the B Pile.

This means that we have (26 - 22) = 4 red cards in the A pile and (26 - 2) = 24 black cards in the A pile.

and 4*6 = 24, which is consistent with the equations that we had before.