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02 Jul 2021, 07:15
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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:
95%
(hard)
Question Stats:
42% (02:06) correct
58%
(01:53)
wrong
based on 254
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive four-digit integers contain the digit grouping “62” (in that order) at least once? For instance,2628 and 6244 are two such integers to include, but 2268 and 5602 do not meet the restrictions.
(A) 180
(B) 190
(C) 279
(D) 280
(E) 360
(A) 180
(B) 190
(C) 279
(D) 280
(E) 360
StrawberryDembaBa
Joined: 05 Jan 2021
Last visit: 14 Mar 2025
Posts: 34
Given Kudos: 20
Location: India
Schools: Anderson '26 (WL) ISB '27 (D) Tepper '26 (WL) Cornell '26 (WL) Foster (A$$) Marshall '26 (A$) NUS '26 (D) Scheller '26 (D) Kenan-Flagler '26 (A$$$) Fuqua '26 (S) Ross '26 (D) INSEAD Aug'24 intake (D)
GMAT 1: 730 Q49 V41
GPA: 3
Schools: Anderson '26 (WL) ISB '27 (D) Tepper '26 (WL) Cornell '26 (WL) Foster (A$$) Marshall '26 (A$) NUS '26 (D) Scheller '26 (D) Kenan-Flagler '26 (A$$$) Fuqua '26 (S) Ross '26 (D) INSEAD Aug'24 intake (D)
GMAT 1: 730 Q49 V41
Posts: 34
02 Jul 2021, 07:40
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Since it has to be a 4-digit number, the possible cases are:
1) 62_ _
Each of the two blanks can take any of the digits from 0-9, hence number of possible cases = 10 * 10 = 100
2) _ 62 _
The first can take any digit except zero, the second blank can take any digit. No. of possible cases = 9*10 = 90
3) _ _ 62
The first can take any digit except zero, the second blank can take any digit. No. of possible cases = 9*10 = 90
But the number 6262 is counted in both the cases 1) and 3). We have to count it only once.
Total cases = 100 +90 +90 -1 = 279
Hence C) is my answer.
1) 62_ _
Each of the two blanks can take any of the digits from 0-9, hence number of possible cases = 10 * 10 = 100
2) _ 62 _
The first can take any digit except zero, the second blank can take any digit. No. of possible cases = 9*10 = 90
3) _ _ 62
The first can take any digit except zero, the second blank can take any digit. No. of possible cases = 9*10 = 90
But the number 6262 is counted in both the cases 1) and 3). We have to count it only once.
Total cases = 100 +90 +90 -1 = 279
Hence C) is my answer.
General Discussion
04 Jul 2021, 05:31
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StrawberryDembaBa
HI,
since the question says "at least once" the answer should be option D that is 280
