If x represents the sum of all the positive three-digit : PS Archive
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# If x represents the sum of all the positive three-digit

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If x represents the sum of all the positive three-digit [#permalink]

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03 Aug 2007, 01:56
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If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222
VP
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03 Aug 2007, 07:59
vshaunak@gmail.com wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I think E.
6 ways to rearrange a,c,b; thus, we have the sum of...
x = abc + acb + bac + bca + cab + cba
x = 100*(2a+2b+2c) + 10*(2a+2b+2c) + (2a+2b+2c)
x = (2a+2b+2c) * (100+10+1)
x = 2 * 111 * (a+b+c)
x = 222 * (a+b+c)
This means that x must be divisible by 222
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03 Aug 2007, 08:49
wow..i wouldnt even know how to read this type of question..i have no clue what it was asking...

good work BKK145..

bkk145 wrote:
vshaunak@gmail.com wrote:
If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I think E.
6 ways to rearrange a,c,b; thus, we have the sum of...
x = abc + acb + bac + bca + cab + cba
x = 100*(2a+2b+2c) + 10*(2a+2b+2c) + (2a+2b+2c)
x = (2a+2b+2c) * (100+10+1)
x = 2 * 111 * (a+b+c)
x = 222 * (a+b+c)
This means that x must be divisible by 222
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03 Aug 2007, 09:15
bkk..i must admit ..that was awsome..

i did solved it my way ..but that was lot clumsier and it too 8 mins to do that...
VP
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03 Aug 2007, 09:20
Thanks guys...encouraging for me. I am definitely stronger in Math, but not so much on verbal

My Q scores vary between 49-51 on GMAT prep. However, took a PR CAT online yesterday and was a blow to my confidence...got Q 44! I don't know why. What you guys said help me boost my confidence up a bit. Much appreciated.
Senior Manager
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03 Aug 2007, 10:54
I did it another way. Just assumed a,b,c to be 1,2,3 respectively. They can be arranged in 3! ways which is 6. In the 6 ways there are 2 1s, 2 2s and 2 3s in the units tens and hundereds digits. Now adding is easy as you get a total of 12. But u need to add 1 when considering the hunderds and tens digits. so the total is 1332. Now this number is divisble by 3, 6, not by 11 and 22 but by 222. so 222 has to be the answer
03 Aug 2007, 10:54
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