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Kaplan: Sum of integers

Expert replies
by shubhamkumar » Sun Apr 15, 2012 12:12 pm
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:C

This one is excellent!
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Source: — Problem Solving |

by shubhamkumar » Sun Apr 15, 2012 12:19 pm
Soln:

Let ABC= 100A+10B+C
BCA= 100B+10C+A
CAB= 100C+10A+B
Sum of the integers in the List= 111(A+B+C)
Factors of 111=1,3,37 and 111
Sum of these Factors = 152
Last edited by shubhamkumar on Wed Apr 18, 2012 10:03 am, edited 1 time in total.
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by Anurag@Gurome » Sun Apr 15, 2012 5:17 pm
shubhamkumar 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:C

This one is excellent!
Sum of three given numbers = (100A + 10B + C) + (100B + 10C + A) + (100C + 10A + B) = 100(A + B + C) + 10(A + B + C) + (A + B + C) = 111(A + B + C)

Now 111 = 37 * 3 implies that the sum should have 1, 3, 37, and 111 as its factors.
So, 1 + 3 + 37 + 111 = 152

The correct answer is C.
Anurag Mairal, Ph.D., MBA
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Gurome, Inc.
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