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If x represents the sum of all the positive three-digit [#permalink]
 13 Nov 2009, 20:35
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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
Last edited by Bunuel on 30 Jul 2012, 04:29, edited 2 times in total.
Added the OA
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Re: this is what it has come down to [#permalink]
13 Nov 2009, 21:34
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rvthryet 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 have never really understood the thinking behind this... Using THREE non-zero digits a,b,c only, we can construct 3!=6 numbers: abc, acb, bac, bca, cab, cba. Their sum will be: x=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b)+(100c+10b+a)==200*(a+b+c)+20*(a+b+c)+2*(a+b+c)==222*(a+b+c)Largest integer by which x MUST be divisible is 222. Answer: E (222).
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Re: Testing number properties [#permalink]
03 Mar 2010, 04:26
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E
Maybe there is a faster way to do it but I did it like this:
How many ways can you arrange abc?
abc
acb
bac
bca
cab
cba
which are equivalent to:
100a + 10b + c
100a + 10c + b
100b + 10a + c
100b + 10c + a
100c + 10a + b
100c + 10b + a
if you add them all together you get 222a + 222b + 222c
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Re: this is what it has come down to [#permalink]
13 Nov 2009, 20:40
where did this question come from wow I have like no idea where to begin I would assume 123 and 987 which are two combinations are both both divisible by 3 as the GCD so 3?
A?
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Re: this is what it has come down to [#permalink]
13 Nov 2009, 21:47
Bunuel wrote: rvthryet 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 have never really understood the thinking behind this... Using THREE non-zero digits a,b,c only, we can construct 3!=6 numbers: abc, acb, bac, bca, cab, cba. Their sum would be: x=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b)+(100c+10b+a)==200*(a+b+c)+20*(a+b+c)+2*(a+b+c)==222*(a+b+c)Largest integer by which x MUST be divisible is 222. Answer: E (222). Good explanation, exactly how I solved it. I love questions with elegant solutions like this. +1
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Re: this is what it has come down to [#permalink]
13 Nov 2009, 21:59
We can also solve this one without math using symmetry: hundreds, tens and units are symmetric, so sum can be written as (y)*111. We need to check that y is even. For example, for fixed a at hundred position, there is two bc,cb combinations. Therefore, a is included twice (even number of times) into sum of hundreds. So, it is 222 By the way, it is the first time when I add something after Bunuel 
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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 I always struggled to understand the theory given in books to solve these types of questions. Can someone please simplify the concept behind solving these questions?
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Re: Largest Integer [#permalink]
31 Jan 2012, 19:03
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Re: If x represents the sum of all the positive three-digit [#permalink]
31 Mar 2012, 03:27
To Bunuel, I've gone thorugh ur notes for each Quant topic and I try to solve topic wise questions from gmatclub. Sometimes I'm not able to figure out how to start with the problem, or I should say how to apply the properties learned since, the techniques you give in your solution for a given problem are not there in properties or formulaes. What do you recommend ? I plan to give my Gmat nxt mnth end. This Tuesday, Veritas prep test I took I scored 600, Q44, verbal 33. Kindly assist. Thanks.
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Re: If x represents the sum of all the positive three-digit [#permalink]
30 Jul 2012, 04:26
.Though I was able to solve it (in a random way), but was unable to come up with a concrete approach.
@NickK kudos for that perfect one. This is how I did.....
The question asked for the largest divisor and thus we need to form 6 largest number that could be made using 3 distinct nonzero digits....987+978+897+879+798+789 = 5328...start from the largest number provided in the answer..222 divides 5328 completely hence is the answer
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Re: If x represents the sum of all the positive three-digit
[#permalink]
30 Jul 2012, 04:26
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