Answer: There are a total of 22 CARDS in pile B
Step-by-step explanation:
It is already certain that the total number of cards that are in pile "A" must be a multiple of 5 due to the fact that there are four times as many black cards as red cards.
If there are 26 red cards and 26 black cards altogether in the piles and each pile must have at least one card, then let's look at the possibilities:
4 black cards, 1 red card in pile A, it will then result to 22 black cards and 25 red cards in pile B. Recall that the number of red cards in pile B must be a multiple of the number of cards in pile B
We check. 25/22 .... (Doesn't agree to multiple rule)
We try another way, 8 black cards, 2 red cards in pile A which then results to 18 black cards and 24 red cards in pile B
(24/18) doesn't agree.
Again 12 black cards, 3 red cards in A, 14 black cards and 23 red cards in pile B... doesn't agree
16 black cards, 4 red cards in pile A, 10 black cards and 22 red cards in pile B... doesn't agree
20 black cards, 5 red cards in pile A, 6 black cards and 21 red cards in pile B... doesn't agree
24 black cards, 6 red cards in pile A, 2 black cards and 20 red cards in pile B... We check if the number of red cards in pile B is a multiple of the number of black cards in pile B:
20/2 = 10 (it agrees to the multiple rule).
So the number of cards in pile B = (number of red cards there + number of black cards there)
= 20 + 2
= 22 cards are in pile B altogether
is symmetric if 
. We have
is indeed symmetric.