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Updated on: 05 Aug 2015, 10:51
Originally posted by GMATinsight on 04 Aug 2015, 02:58.
Last edited by GMATinsight on 05 Aug 2015, 10:51, edited 3 times in total.
Last edited by GMATinsight on 05 Aug 2015, 10:51, edited 3 times in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
34% (02:38) correct
66%
(02:49)
wrong
based on 191
sessions
History
Date
Time
Result
Not Attempted Yet
Q-4: How many three digit numbers of distinct digits can be formed by using digits 1, 2, 3, 4, 5, 6 and 7 such that the numbers are divisible by 11?
A) 14
B) 16
C) 18
D) 20
E) 22
Source : https://www.GMATinsight.com
Good Questions also deserve Kudos :)
A) 14
B) 16
C) 18
D) 20
E) 22
Source : https://www.GMATinsight.com
Good Questions also deserve Kudos :)
05 Aug 2015, 10:49
Kudos
Bookmarks
GMATinsight
Q-4: How many three digit numbers of distinct digits can be formed by using digits 1, 2, 3, 4, 5, 6 and 7 such that the numbers are divisible by 11?
A) 14
B) 16
C) 18
D) 20
E) 22
Source : https://www.GMATinsight.com
Good Questions also deserve Kudos :)
A) 14
B) 16
C) 18
D) 20
E) 22
Source : https://www.GMATinsight.com
Good Questions also deserve Kudos :)
CONCEPT: A number is divisible by 11 when Difference of Sums of even place digits and odd place digits is either zero or a multiple of 11.
i.e. a 3-Digit Number abc will be divisible by 11 if (a+c) - b = 0 or multiple of 11
@b=1, a+c should be 1 or 12 i.e. (a,c) can be (5, 7), (7, 5) --- 2 cases
@b=2, a+c should be 2 or 13 i.e. (a,c) can be (6, 7), (7, 6) ---- 2 cases
@b=3, a+c should be 3 or 14 i.e. (a,c) can be (1, 2), (2, 1) ---- 2 cases
@b=4, a+c should be 4 or 15 i.e. (a,c) can be (1, 3), (3, 1) ---- 2 cases
@b=5, a+c should be 5 or 16 i.e. (a,c) can be (1, 4), (2, 3), (3, 2), (4, 1) ---- 4 cases
@b=6, a+c should be 6 or 17 i.e. (a,c) can be (1, 5), (2, 4), (4, 2), (5, 1) ---- 4 cases
@b=7, a+c should be 7 or 18 i.e. (a,c) can be (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) ---- 6 cases
Total Cases = 2+2+2+2+4+4+6 = 22 cases
Answer: option E
General Discussion
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,333
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
Updated on: 05 Aug 2015, 11:33
Originally posted by ENGRTOMBA2018 on 04 Aug 2015, 04:35.
Last edited by ENGRTOMBA2018 on 05 Aug 2015, 11:33, edited 4 times in total.
Last edited by ENGRTOMBA2018 on 05 Aug 2015, 11:33, edited 4 times in total.
Updated the possible cases from 18 to 22.
Kudos
Bookmarks
GMATinsight
Q-4: How many three digit numbers of distinct digits can be formed by using digits 1, 2, 3, 4, 5, 6 and 7 such that the numbers are divisible by 11?
A) 8
B) 10
C) 12
D) 14
E) 16
Source : https://www.GMATinsight.com
Good Questions also deserve Kudos :)
A) 8
B) 10
C) 12
D) 14
E) 16
Source : https://www.GMATinsight.com
Good Questions also deserve Kudos :)
Did by counting
132
143
154
165
176
231
253
264
275
341
352
374
451
462
473
517
561
572
627
671
715
726
GMATinsight, Total should be 22.
