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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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C
D
E

Difficulty:

  75% (hard)

Question Stats:

45% (03:29) correct 55% (02:48) wrong based on 20 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
1
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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Re: Four friends have some coins. A has 2 less than B, who has 2 less than   [#permalink] 07 Apr 2019, 02:25
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