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If x represents the sum of all the positive threedigit
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Updated on: 30 Jul 2012, 04:29
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If x represents the sum of all the positive threedigit 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
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Originally posted by rvthryet on 13 Nov 2009, 20:35.
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
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13 Nov 2009, 21:34
rvthryet wrote: If x represents the sum of all the positive threedigit 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 nonzero 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: this is what it has come down to
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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
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13 Nov 2009, 21:47
Bunuel wrote: rvthryet wrote: If x represents the sum of all the positive threedigit 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 nonzero 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
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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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Re: Testing number properties
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03 Mar 2010, 04:26
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: If x represents the sum of all the positive threedigit
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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 threedigit
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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 threedigit
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01 Oct 2013, 06:40
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.
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Re: this is what it has come down to
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02 Oct 2013, 00:27
Bunuel wrote: rvthryet wrote: If x represents the sum of all the positive threedigit 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 nonzero 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). 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!



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Re: this is what it has come down to
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02 Oct 2013, 03:12
shameekv wrote: Bunuel wrote: rvthryet wrote: If x represents the sum of all the positive threedigit 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 nonzero 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). 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). Hope it helps.
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Re: If x represents the sum of all the positive threedigit
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02 Oct 2013, 03:27
Hi Bunuel,
Thanks for the clarification. I thought it is the sum of all such 3digit numbers that have distinct numbers.
What in the case "x is the sum of all the 3digit 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?
Thanks, Shameek



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Re: If x represents the sum of all the positive threedigit
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26 Oct 2014, 12:52
Shamee, to solve the problem in a simpler manner why don't you assume the numbers a, b and c to be 1, 2 and 3 respectively?
Thus, the distinct numbers that can be formed would be  123 132 213 231 312 321
If you sum these up you get a total of 1332.
Then proceed to plug in the answer options to find the greatest number that divides 1332.
From the options  (A) 3  Yes (B) 6  Yes (C) 11  No (D) 22  No (E) 222  Yes
Clearly, since 222 is the greatest, E is the right option.



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Re: If x represents the sum of all the positive threedigit
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18 Jan 2016, 22:29
pritishpratap wrote: Shamee, to solve the problem in a simpler manner why don't you assume the numbers a, b and c to be 1, 2 and 3 respectively?
Thus, the distinct numbers that can be formed would be  123 132 213 231 312 321
If you sum these up you get a total of 1332.
Then proceed to plug in the answer options to find the greatest number that divides 1332.
From the options  (A) 3  Yes (B) 6  Yes (C) 11  No (D) 22  No (E) 222  Yes
Clearly, since 222 is the greatest, E is the right option. Here is the catch in "assuming values" in this question: The question is a "must be true" question. How do you know that what holds for values 1, 2 and 3 will be true for values say 2, 3 and 7 too? What if sum of numbers formed by 2, 3 and 7 is not divisible by 222? You do need to apply logic to confirm "must be true".
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Re: If x represents the sum of all the positive threedigit
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31 Mar 2017, 00:56
rvthryet wrote: If x represents the sum of all the positive threedigit 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 Bunuel, This question has been wrongly tagged. The original source is Manhattan Prep, Challenge Problems (2002, December 2, ThreeDigit Divisibility).
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Re: If x represents the sum of all the positive threedigit
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31 Mar 2017, 01:50
ziyuen wrote: rvthryet wrote: If x represents the sum of all the positive threedigit 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 Bunuel, This question has been wrongly tagged. The original source is Manhattan Prep, Challenge Problems (2002, December 2, ThreeDigit Divisibility). Edited. Thank you.
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If x represents the sum of all the positive threedigit
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09 Oct 2017, 03:10
I solved it in a bit different way. Not sure if this is correct. There are 3 numbers a, b, c so there can be 6 arrangements of these numbers. So, 6 possible numbers are there ( just to be sure i am not missing) abc +acb +bac +bca +cba +cab6a+6b+6cNow if we factor out 6 > 6( a+b+c ) from this we know the answer must be a multiple of 6. It can not be 6 as a+b+c wpuld yield some integer and 6*someinteger > 6 . So the only possible outcome is 222 which is a multiple of 6 other than Choice B. Answer E
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Re: If x represents the sum of all the positive threedigit
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22 Aug 2018, 06:30
Bunuel wrote: rvthryet wrote: If x represents the sum of all the positive threedigit 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 nonzero 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). I tried solving it using the formula (n1)!*(sum of the digits)*(111…..n times) n =3 sum of the digits = a+b+c so (31)! *(a+b+c)*111 222*(a+b+c) Hence clearly the number will be divisible by 222



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Re: If x represents the sum of all the positive threedigit
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06 Sep 2019, 07:46
In my humble opinion, although this is not exactly a problem that is tested extensively on the GMAT, there’s no harm in solving questions like these to strengthen your concepts of Permutations with numbers. If a, b and c are the digits from which we have to form 3digit numbers such that all the digits are nonzero and distinct, then, the number of ways of doing that is given by, (a+b+c) * (31)! * (111). In general, if there are ‘n’ distinct digits using which we have to form ‘n’ digit numbers where the digits are nonzero and distinct, then the number of ways of doing this is given by, (Sum of the n digits) *(n1)! *(1111….. n times). The reason for this is as follows: When a comes in the units place, its place value will be a. There will be (n1)! numbers where a will be the units digit. Therefore, the sum of these values will be a * (n1)! When a comes in the tens place, its value will be 10a. There will be (n1)! numbers where a will be the tens digit. The sum of these values will be 10a * (n1)! When a comes in the hundreds place, its value will be 100a. There will be (n1)! numbers where a will be the hundreds digit. The sum of these values will be 100a * (n1)!. The sum of all these values will be (n1)! * a (100 + 10 + 1) = (n1)! * a * 111. Similarly, for the other digits, b, c, d and so on, the sum of the values can be worked out as shown above. In our case, since the sum total is (a+b+c) * 2! * 111 which is nothing but (a+b+c) * 222, we can say that the sum will be definitely divisible by 222. The correct answer option is E. Hope this helps!
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