Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
18 Jan 2022, 03:30
Kudos
Bookmarks
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:
38% (02:31) correct
62%
(02:19)
wrong
based on 82
sessions
History
Date
Time
Result
Not Attempted Yet
Let abcd be a general four-digit number and all the digits are non-zero. How many four-digits numbers abcd exist such that the four digits are all distinct and such that a + b + c = d ?
(A) 6
(B) 7
(C) 24
(D) 36
(E) 42
Are You Up For the Challenge: 700 Level Questions
(A) 6
(B) 7
(C) 24
(D) 36
(E) 42
Are You Up For the Challenge: 700 Level Questions
18 Jan 2022, 05:35
Kudos
Bookmarks
Smallest possible number possible for abcd where a+b+c = d is 6 and the greatest is 6.
1+2+x:
smallest value x can be is 3, while the greatest is 6.
\(6-3+1=4\) There are 4 numbers for x which when added to 1 and 2, will equal to a digit.
4 NUMBERS
1+3+x:
x can only be 4 or 5. 2 is already covered in the previous part. if x is 4 we get 8 and if x is 5 we get 9.
2 NUMBERS
2+3+x:
2+3 = 5, which means 1 ≤ x ≤ 4. However, 1 has already been covered, 2 and 3 won't work as they would not be distinct, which leaves only 4.
1 NUMBER
Total: \(4+2+1=7\) so there are 7 total numbers which work. However, the 3 digits before d (abc), can be arranged in \(3!\) ways.
\(7*3! = 42\)
Answer E
1+2+x:
smallest value x can be is 3, while the greatest is 6.
\(6-3+1=4\) There are 4 numbers for x which when added to 1 and 2, will equal to a digit.
4 NUMBERS
1+3+x:
x can only be 4 or 5. 2 is already covered in the previous part. if x is 4 we get 8 and if x is 5 we get 9.
2 NUMBERS
2+3+x:
2+3 = 5, which means 1 ≤ x ≤ 4. However, 1 has already been covered, 2 and 3 won't work as they would not be distinct, which leaves only 4.
1 NUMBER
Total: \(4+2+1=7\) so there are 7 total numbers which work. However, the 3 digits before d (abc), can be arranged in \(3!\) ways.
\(7*3! = 42\)
Answer E
20 Sep 2025, 05:25
Kudos
Bookmarks
Bunuel is there a better approach for this?
I had to make the subsets...
d has to be atleast 6...
So 6 can be achieved in ....1 + 2 + 3 ...Can arrange this in 3! ways..
if d is 7..Then 1 + 2 + 4 ...3! ways again ....
d is 8...then 1 + 2 + 5 ..or 1 + 3 + 4 ....each 3! ways..
d is 9 ...then 1 + 2 + 6 or 1 + 5 + 3 or 2 + 3 + 4 ..all in 3! ways..Answer comes to 42 ...but I took a lot of time and I had no confidence..as i thought there could be some combination that I am not able to get...Is there a way to find that out?..What different digits when added together gives a particular different digit? or we have to do that manually, like the way I did...?
I had to make the subsets...
d has to be atleast 6...
So 6 can be achieved in ....1 + 2 + 3 ...Can arrange this in 3! ways..
if d is 7..Then 1 + 2 + 4 ...3! ways again ....
d is 8...then 1 + 2 + 5 ..or 1 + 3 + 4 ....each 3! ways..
d is 9 ...then 1 + 2 + 6 or 1 + 5 + 3 or 2 + 3 + 4 ..all in 3! ways..Answer comes to 42 ...but I took a lot of time and I had no confidence..as i thought there could be some combination that I am not able to get...Is there a way to find that out?..What different digits when added together gives a particular different digit? or we have to do that manually, like the way I did...?
Bunuel
