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

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Four friends have some coins. A has 2 less than B, who has 2 less than  [#permalink]

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New post 11 Nov 2010, 14:10
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A
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E

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Question Stats:

60% (03:12) correct 40% (03:00) wrong based on 42 sessions

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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
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Re: Four friends have some coins. A has 2 less than B, who has 2 less than  [#permalink]

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New post 11 Nov 2010, 14:25
1
cleetus wrote:
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


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 4 consecutive even integers (they cannot be odd as given that their product is even 5760) --> \((A-2)A(A+2)(A+4)=5760=6*8*10*12\) (after some trial and error) --> \(6+8+10+12=36\).

Answer: E.
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Re: Four friends have some coins. A has 2 less than B, who has 2 less than  [#permalink]

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New post 11 Nov 2010, 14:25
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1
cleetus wrote:
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


Let the number of coins be a,b,c,d

b=c-2
a=b-2=c-4
d=a-2=c-6

a*b*c*d=5760
c(c-2)(c-4)(c-6)=5760

Now since the product is even, it is obvious c must be even, let c=2x

2x(2x-2)(2x-4)(2x-6)=5760
x(x-1)(x-2)(x-3)=360

x is an integer, and x must be greater than or equal to 4, so should be very easy to check
x=4 ... product is 24
x=5 ... product is 120
x=6 ... product is 360 : Bingo !!

Hence the solution is c=12, b=10, a=8, d=6

Sum = 36 --> Answer : e
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Re: Four friends have some coins. A has 2 less than B, who has 2 less than  [#permalink]

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New post 11 Nov 2010, 14:31
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
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Re: Four friends have some coins. A has 2 less than B, who has 2 less than  [#permalink]

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New post 11 Nov 2010, 17:18
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.
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Re: Four friends have some coins. A has 2 less than B, who has 2 less than  [#permalink]

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New post 11 Nov 2010, 19:29
cleetus wrote:
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


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.
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Re: Four friends have some coins. A has 2 less than B, who has 2 less than  [#permalink]

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New post 23 Jun 2019, 11:28
cleetus wrote:
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


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
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Re: Four friends have some coins. A has 2 less than B, who has 2 less than  [#permalink]

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New post 23 Jun 2019, 21:35
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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Re: Four friends have some coins. A has 2 less than B, who has 2 less than   [#permalink] 23 Jun 2019, 21:35
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