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If x represents the sum of all the positive three-digit [#permalink]
13 Nov 2009, 19:35

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Question Stats:

45% (03:20) correct
55% (02:05) wrong based on 652 sessions

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?

Re: this is what it has come down to [#permalink]
13 Nov 2009, 20:34

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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...

Re: If x represents the sum of all the positive three-digit [#permalink]
19 Sep 2013, 09:20

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Hello from the GMAT Club BumpBot!

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Re: this is what it has come down to [#permalink]
13 Nov 2009, 19:40

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?

Re: this is what it has come down to [#permalink]
13 Nov 2009, 20:47

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...

Re: this is what it has come down to [#permalink]
13 Nov 2009, 20:59

Expert's post

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 _________________

Re: If x represents the sum of all the positive three-digit [#permalink]
31 Mar 2012, 02:27

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.

Re: If x represents the sum of all the positive three-digit [#permalink]
30 Jul 2012, 03:26

.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

Re: If x represents the sum of all the positive three-digit [#permalink]
01 Oct 2013, 05:40

ratinarace wrote:

.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

Agree, substitution works the best for 'must be true' problems. _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: this is what it has come down to [#permalink]
01 Oct 2013, 23:27

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...

Re: this is what it has come down to [#permalink]
02 Oct 2013, 02:12

Expert's post

shameekv wrote:

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...

Largest integer by which x MUST be divisible is \(222\).

Answer: E (222).

Hi Bunuel, Can you please explain me what will be the value of "x" in this question. If it were asked what is the value of x?

Thanks!

We cannot say what x is.

If a, b, and c, are 1, 2, and 3 respectively, then x = 123 + 132 + 213 + 231 + 312 + 321 = 1,332 = 6*222 (the least possible value of x). ... If a, b, and c, are 7, 8, and 9 respectively, then x = 789 + 798 + 879 + 897 + 978 + 987 = 5,328 = 24*222 (the greatest possible value of x).

Re: If x represents the sum of all the positive three-digit [#permalink]
02 Oct 2013, 02:27

Hi Bunuel,

Thanks for the clarification. I thought it is the sum of all such 3-digit numbers that have distinct numbers.

What in the case "x is the sum of all the 3-digit numbers that have distinct numbers". How do you calculate the value of x in such case. I tried many things but couldn't work it out.

I saw such type of question recently where x was required to be calculated but the digits could be repeated and that made it simple. But I couldn't figure out with this restriction. Could you please help me out on that?

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