Four friends have some coins

[BTGmoderatorRO]

Four friends have some coins. A has 2 less than B, who has 2 less than C. If D has 2 less than A and the product of the number of coins that each of them has is 5760, how many coins in all do they have among themselves?

A) 38
B) 40
C) 42
D) 46
E) 36

OA is E
This question looks tricky, how do I approach it? I need an answer from an expert. Thanks in anticipation.

[Scott@TargetTestPrep]

Four friends have some coins. A has 2 less than B, who has 2 less than C. If D has 2 less than A and the product of the number of coins that each of them has is 5760, how many coins in all do they have among themselves?

A) 38
B) 40
C) 42
D) 46
E) 36

OA is E

We can let the number of coins D has be n, so A, B, and C have (n + 2), (n + 4), and (n + 6) coins, respectively. Despite this, each person actually has about the same number of coins. So we are looking for a number such that its fourth power is about 5760. That is, let's take the fourth root of 5760, which is about 8.7. Let's round it down to 8 since the number of coins has to be a whole number.

Now let's say D, A, B, and C have 8, 10, 12, and 14 coins, respectively. However, the product 8 x 10 x 12 x 14 = 13,440 is not 5760 we are looking for.

Notice that the number of coins each one has must be an even number. That is because n, n + 2, n + 4 and n + 6 must be either all odd or all even. However, since their product is even, it means they must be all even.

So let's say D, A, B and C have 6, 8, 10, and 12 coins, respectively. Let's check the product again:

6 x 8 x 10 x 12 = 5760

Since it's exactly 5760, we have found the number of coins each one has, and therefore, the total number of coins they have is 6 + 8 + 10 + 12 = 36.

Answer: E