Find all possible values of X - advanced question - The Beat The GMAT Forum

Find all possible values of X - advanced question

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adgir: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sun Jan 08, 2012 6:52 pm

by adgir » Sun Jan 08, 2012 7:01 pm

X is a 3-digit number with all digits being distinct.

There are 9 2-digit numbers that can be created using the digits of X.

X is a sum of all 9 of these 2-digit numbers.

Find all possible values of X.

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Source: Beat The GMAT — Problem Solving | Sun Jan 08, 2012 7:01 pm

shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Mon Jan 09, 2012 12:46 am

adgir wrote:
Find all possible values of X.

0 cannot be one of the digits because 9 2 digit integers cannot be formed.

leaving 0 out, we still have 9 digits and 9C3 = 84 possible 3 digit integers combinations

who in her/his sane mind would calculate that?

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ronnie1985: Legendary Member; Posts: 626; Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members; Location: Ahmedabad; Thanked: 31 times

by ronnie1985 » Mon Jan 09, 2012 12:56 am

XYZ is the no. , say.
Then 100X+10Y+Z = 3(X+Y+Z)+30(X+Y+Z) = 33(X+Y+Z)
Therefore, the no of integral solutions of the equation 23Y+32Z = 67,134,201,268,335,402,469,536,603.
Which can be obtained by getting the coeff of x^(RHS of Equation) in the product: (1+x^23+x^46+...)*(1+x^32+x^64+...)
Quite lengthy

Follow your passion, Success as perceived by others shall follow you

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