Four friends have some coins. A has 2 less than B, who has 2 less than

Suppose C has c coins. So
C=c
B= c-2
A= b-2= (c-2)-2 = c-4
D= a-2= (c-4)-2 = c-6

Sum of all coins
= c+(c-2)+(c-4)+(c-6)
=4c-12

Now
c*(c-2)*(c-4)*(c-6)= 5760
c*(c-2)*(c-4)*(c-6)=576*10
=24*24*10
=12*10*8*6 (Re-adjusting as per c multiples )

Therefore c= 12 & Total coins=4c-12=36

Answer:- E

Just like everyone else,

I got

\(c (c - 2) (c - 4) (c - 6) = 5760\)

I simplified 5760 info prime factors and got \(2^7*3^2*5\)

since the product is even I know I had to some how get the 2 3s and 5 into even by multiplying them by the abundant supply of 2s. And I also knew that the 4 numbers had to be consecutive even because 5760 demands even and c (c - 2) (c - 4) (c - 6) demands consecutive numbers.

So I multiplied 5*2 = 10 and 3*2 to get 6. To fill in the gap, 2^3 = 8 and 2^2 * 3 = 12.

So 6,8,10,12 fits all the criteria.

I think the above solutions are all pretty good but I figured it using another approach so I am writing it down.

Seeing that each friend has 2 less than another friend, I figured they are even consecutive integers (even because the product is even).

Then I thought the product is 5760 (which is 24 * 24 * 10 because 24^2 = 576)
Then I thought of the number which, to the power 4 is around 5760. because the required even numbers would be around this number.

10^4 = 10,000. So the number has to be less than 10, I said.
Next number I thought of was 8 since 8 = 2^3 and I know 2^12 = 4096 which is a little less than 5760.
Therefore, I suspected 6 * 8 * 10 * 12 will work out which I confirmed by cross checking with 24 * 24 * 10. Add them to get 36.

Some brute force, some logic but you get your answer in 30 secs.

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

Given # of coins:
A;
B=A+2;
C=B+2=A+4;
D=A-2;

So # of coins are: A-2, A, A+2, A+4, so total = 4A+4
And only 36 can be equal to 4A+4 , thus answer is E

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