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alphonsa
Joined: 22 Jul 2014
Last visit: 25 Oct 2020
Posts: 106
Given Kudos: 197
Concentration: General Management, Finance
Schools: Cox '19 (D) Babson '19 (A$) Katz '19 (D) Eller"19 (A$) Northeastern '19 (A$) Krannert '19 (A$) ISB '18 (D) Rutgers '19 (A) Olin '19 (I)
GMAT 1: 670 Q48 V34
WE:Engineering (Energy)
Products:
Schools: Cox '19 (D) Babson '19 (A$) Katz '19 (D) Eller"19 (A$) Northeastern '19 (A$) Krannert '19 (A$) ISB '18 (D) Rutgers '19 (A) Olin '19 (I)
GMAT 1: 670 Q48 V34
Posts: 106
03 Aug 2014, 09:49
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
78% (01:40) correct
22%
(01:43)
wrong
based on 225
sessions
History
Date
Time
Result
Not Attempted Yet
What is the least positive integer that when divided by 3,6,9 leaves remainders of 2 in each case but is perfectly divisible by 11?
A) 33
B) 77
C)110
D) 121
Besides plugging in, can you tell me if there is another method of solving this problem? I mean just for a conceptual understanding.
Source: 4gmat
A) 33
B) 77
C)110
D) 121
Besides plugging in, can you tell me if there is another method of solving this problem? I mean just for a conceptual understanding.
Source: 4gmat
03 Aug 2014, 15:34
Kudos
Bookmarks
3=3*1, 6=3*2, 9=3*3,
so, Least Common Multiple [3,6,9] = 3*3*2=18
Let A=18X+2 = 11Y, when X =6, Y= 10, A=110
so, Least Common Multiple [3,6,9] = 3*3*2=18
Let A=18X+2 = 11Y, when X =6, Y= 10, A=110
03 Aug 2014, 23:56
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alphonsa
I think the fastest way is try and error. I use the tips that try from E to A, instead of A to E, will save more time.
I got C
