Convolated a hell!

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pareekbharat86: Master | Next Rank: 500 Posts; Posts: 168; Joined: Thu Nov 01, 2012 7:43 pm; Thanked: 1 times

by pareekbharat86 » Sun Nov 10, 2013 2:40 am

List L: ABC, BCA, CAB

In list L above, there are 3 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?

a. 47
b. 114
c. 152
d. 161
e. 488

OA is C.

This one must be for the creme de la creme.

Source- Kaplan Practice Test 3

Thanks,
Bharat.

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Source: Beat The GMAT — Problem Solving | Sun Nov 10, 2013 2:40 am

Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Sun Nov 10, 2013 3:35 am

pareekbharat86 wrote:List L: ABC, BCA, CAB

In list L above, there are 3 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?

a. 47
b. 114
c. 152
d. 161
e. 488

OA is C.

This one must be for the creme de la creme.

Source- Kaplan Practice Test 3

Hi Bharat,
Sum of Integers in the list L = (100A+10B+C) + (100B+10C+A) + (100C+10A+B) = 111*(A + B + C)
So, sum is multiple of 111.
Factors of 111 is 1,3,37(3*37=111)

Therefore factors of 111*(A + B + C) are 1,3,37,111.
Sum of Factors is (1+3+37+111 =152)

Answers is C

Hope it helps you.

Regards,
Uva.

Known is a drop Unknown is an Ocean

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Nov 10, 2013 4:03 am

List L: ABC, BCA, CAB

In list L above, there are 3 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?

47
114
152
161
188

Integer ABC = 100A + 10B + C.
Integer BCA = 100B + 10C + A.
Integer CAB = 100C + 10A + B.
Sum of all 3 integers =

= (100A + 100B + 100C) + (100B + 10C + A) + (100C + 10A + B)

= (100A + 10A + A) + (100B + 10B + B) + (100C + 10C + C)

= 111A + 111B + 111C

= 111(A+B+C).

Thus, every factor of 111 must be a factor of ABC + BCA + CAB.
Factors of 111:
1 * 111
3 * 37.
Sum of all these factors = 1+3+37+111 = 152.

The correct answer is C.

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Milovan: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Sun Feb 24, 2013 2:29 am; Location: Belgrade; Thanked: 1 times

by Milovan » Sun Nov 10, 2013 12:54 pm

Sum of these three integers is equal to 111 * (A+B+C) and that is clear. All factors of 111 are factors of the sum. However, what if sum of A+B+C alone is divisible by 5. In this case 5 would also be a factor of the sum of three integers.

Some comments?

After posting the question I saw word MUST, that means not ALL factors

Cool, now it is clear...

Milovan Arnaut

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Tue Nov 12, 2013 12:16 pm

I'm a bit bemused by the question. To unramble my ambiguous mind, could someone please let me know which of the statements is correct in the question:
A) "the sum of (all the positive integers (that must be factors of the sum of the integers) in the list L))"
B) "the sum of (all the positive integers that must be factors of the sum of (the integers in the list L))"
C) "the sum of (all the positive integers that must be factors of the sum of the integers) in list L"
Furthermore, if B or C apply, does "all the positive integers" refer to any positive integers, or just the limited integers A, B and C?
If C applies, does the "sum of the integers" refer to any integers, the integers earlier mentioned, or A,B,C, or all 9 integers in list L?

If you can get your head around all the above, then you will see why I can't identify the question. It seems to me that it could mean so many things that it is unanswerable in it's current format. Clarity would be much appreciated. Many thanks.

l

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Nov 13, 2013 4:19 am

Mathsbuddy wrote:Clarity would be much appreciated. Many thanks.

Here is a rephrased version of the problem:

S = ABC + BCA + CAB, where A, B and C are 3 different nonzero digits. List L is composed of every positive integer that MUST be a divisor of S. What is the sum of the integers in List L?

Last edited by GMATGuruNY on Wed Nov 13, 2013 4:22 am, edited 1 time in total.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Nov 13, 2013 4:20 am

[duplicate post - deleted]

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Wed Nov 13, 2013 8:43 am

Having churned this one over in my head for a couple of days, I'm still not convinced the question is right.

"In list L above, there are 3 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?"

Underlined above: factors must be, by definition, positive integers, therefore the question could be written simply as:

In list L above, there are 3 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the factors of the sum of the integers in list L?

Eitherway, all the factors of the sum of the integers in list L are:

1, 3, 37, 111 and (A+B+C) as well as any other factors of (A+B+C). If we ignore the latter, the sum of all the factors is still (152 + A + B + C) and as A,B and C are non zero, this means that 152 cannot be the sum of all of them, as required by the question.

Let's look at the rephrased question:
"
S = ABC + BCA + CAB, where A, B and C are 3 different nonzero digits. List L is composed of every positive integer that MUST be a divisor of S. What is the sum of the integers in List L?"

L = 1, 3, 37, 111 and (A+B+C) as well as any different divisors of (A+B+C). If we ignore the latter, the sum of L = (152 + A + B + C) and as A,B and C are non zero, this means that 152 cannot be the sum of all of them, as required by the question.

Both statements concur with both versions of the question. Furthermore, why should the other factors of (A+B+C) be ignored?

I believe that the sum of L = (152 + A + B + C + remaining factors of (A+B+C)) > 152

In fact, the minimum value of ABC = 123, gives A+B+C (minimum) = 6 (with factors 1, 2, 3, 6)

Therefore L (minimum) = (152 + 1 + 2 + 3 + 6)= 164

This eliminates answers (A), (B), (C) and (D).
Therefore (assuming that 188 is valid) the only remaing possible answer is (E).

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Wed Nov 13, 2013 8:57 am

GMATGuruNY wrote:
Mathsbuddy wrote:Clarity would be much appreciated. Many thanks.
Here is a rephrased version of the problem:

S = ABC + BCA + CAB, where A, B and C are 3 different nonzero digits. List L is composed of every positive integer that MUST be a divisor of S. What is the sum of the integers in List L?

Thank you, but please see my next entry.

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Wed Nov 13, 2013 9:04 am

To simplify what I wrote in the last posting:

(A+B+C) is a factor of the sum of the integers in list L

and

(A+B+C) is a positive integer

Therefore (A+B+C) has to be included in "all the positive integers that MUST be factors of the sum of the integers in list L?"

Ditto for any other factors of (A+B+C)

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Nov 13, 2013 10:22 am

Mathsbuddy wrote:To simplify what I wrote in the last posting:

(A+B+C) is a factor of the sum of the integers in list L

and

(A+B+C) is a positive integer

Therefore (A+B+C) has to be included in "all the positive integers that MUST be factors of the sum of the integers in list L?"

Ditto for any other factors of (A+B+C)

As shown in my initial post, S = 111(A+B+C).
List L is composed of every positive integer that MUST be a factor of S.
Must means IN EVERY CASE.

Case 1: A=1, B=2, C=3
Here, S = 111(1+2+3) = 111*7.
Factors of 111*7 are 1, 3, 7, 37, 111.

Case 2: A=1, B=2, C=8
Here, S = 111(1+2+8) = 111*11.
Factors of 111*11 are 1, 3, 11, 37, 111.

Case 3: A=1, B=4, C=8
Here, Here, S = 111(1+4+8) = 111*13.
Factors of 111*13 are 1, 3, 13, 37, 111.

As the cases above illustrate, only the values in red MUST be factors of S.
Since 7 is NOT a factor in Cases 2 and 3, it is not true that 7 MUST be a factor of S.
Since 11 is NOT a factor in Cases 1 and 3, it is not true that 11 MUST be a factor of S.
Since 13 is NOT a factor in Cases 1 and 2, it is not true that 13 MUST be a factor of S.

Result:
L = {1, 3, 37, 111}.
Sum = 1+3+37+111 = 152.

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Thu Nov 14, 2013 1:43 am

GMATGuruNY wrote:
Mathsbuddy wrote:To simplify what I wrote in the last posting:

(A+B+C) is a factor of the sum of the integers in list L

and

(A+B+C) is a positive integer

Therefore (A+B+C) has to be included in "all the positive integers that MUST be factors of the sum of the integers in list L?"

Ditto for any other factors of (A+B+C)
As shown in my initial post, S = 111(A+B+C).
List L is composed of every positive integer that MUST be a factor of S.
Must means IN EVERY CASE.

Case 1: A=1, B=2, C=3
Here, S = 111(1+2+3) = 111*7.
Factors of 111*7 are 1, 3, 7, 37, 111.

Case 2: A=1, B=2, C=8
Here, S = 111(1+2+8) = 111*11.
Factors of 111*11 are 1, 3, 11, 37, 111.

Case 3: A=1, B=4, C=8
Here, Here, S = 111(1+4+8) = 111*13.
Factors of 111*13 are 1, 3, 13, 37, 111.

As the cases above illustrate, only the values in red MUST be factors of S.
Since 7 is NOT a factor in Cases 2 and 3, it is not true that 7 MUST be a factor of S.
Since 11 is NOT a factor in Cases 1 and 3, it is not true that 11 MUST be a factor of S.
Since 13 is NOT a factor in Cases 1 and 2, it is not true that 13 MUST be a factor of S.

Result:
L = {1, 3, 37, 111}.
Sum = 1+3+37+111 = 152.

Thank you very much for this. I have slept on it and now see how "MUST" indicates this restriction; and I understand fully what you mean.

However, I do believe that the "MUST" can still be interpretted 2 different ways.

If the question had begun:

"L is the list of...", then I would agree 100%.
"L is a list of...", then I would not agree, as MUST could apply to that one singular (undetermined) case.

As the question stands, it is not clear whether L covers all cases, or L is just one particular (unspecified) case.

For example, consider the question:

List L: 123, 231, 312

In list L above, there are 3 positive integers, where each digit is different and nonzero. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?

This does not conflict with the given question, therefore the possibility of a individual case cannot be ignored - even with "must" present. Here the "must" could refer to the singular case given, emphasising that the required numbers cannot be anything but factors.

Here, S = 111(1+2+3) = 111*6.
Factors of 111*6 are 1, 3, 6, 37, 111, 666

Nonetheless, in the generic list, I see how the "must" comes into play, but I don't feel that the question is explicit enough in this regard, to make this assumption certain.

Otherwise, it's a great problem which is easy to solve once you understand the wording. I've learned a lot from it, and I'm swayed into your point of you!

Thanks.

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