Distinct Values

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satyavegi: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Sat Aug 13, 2011 1:07 am

by satyavegi » Fri Jun 01, 2012 10:26 am

Q.The product of three consecutive positive integers is three times their sum.One of this integer is sum of other two integers.if the product of this three integers is denoted by X.Then find the sum of all possible distinct values of X.

A)36
b)72
c)84
d)18
e)24

[spoiler]OA)B[/spoiler]

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Source: Beat The GMAT — Problem Solving | Fri Jun 01, 2012 10:26 am

eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Fri Jun 01, 2012 5:45 pm

Hi satyavegi:

Please check the wording of the question, there is no solution to the problem as it is written. The only consecutive integers where one integer is sum of the others are 1,2,and 3. But this does not meet the Product = 3 * Sum condition. Please post the correct question.

Let me know if this helps

Last edited by eagleeye on Sat Jun 02, 2012 3:26 am, edited 1 time in total.

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ankita1709: Master | Next Rank: 500 Posts; Posts: 109; Joined: Wed Feb 15, 2012 7:09 amFollowed by: 2 members; Thanked: 8 times

by ankita1709 » Sat Jun 02, 2012 3:19 am

There can be solution to the problem if consecutive integers are considered.
But, if the integers are not consecutive the following solution can be used

Lets assume 3 integers as a,b,c

Equation 1- a*b*c = 3(a+b+c)
Equation 2- c= a+b

Replacing c in equation 1

a*b(a+b)=3*2(a+b)
We can cancel out a+b as it will be a non-zero value

we get a*b = 6
Possible values of a and b - 2,3 or 6,1

hence c = 5 or 7

X = a*b*c = 30 or 42

Adding both 30+42= 72

Hence the solution is B

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