Events & Promotions
| Last visit was: 06 Sep 2026, 18:20 |
It is currently 06 Sep 2026, 18:20 |
|
|
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.
Sep 0712:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT- Sep 07
10:00 AM PDT
-11:00 AM PDT
A complete walkthrough of GMAT Club’s free 12-week GMAT study plan: what you study each week, how progress and goals are tracked, how the error log works, and how your practice unlocks paid tools at no cost. 
Sep 0201:00 PM EDT
-11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 0410:00 AM EDT
-11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway - Sep 09
10:00 AM PDT
-11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application... 
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2108:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
11 Jul 2026, 06:06
Kudos
Bookmarks
It is given in the question that AB + BA = CDC
And we are asked the unique number of pair of A*B*C*D
This means we need to find out the total possible pairs of A, B , C and D and then the unique pairs
Now - since we are reversing the 2 digits, and the total becomes a 3 digit number, then the possible pairs can be any pairs starting from 90 to 99 except for 90 and 91 since it leads to 0 in answer and in the question it is mentioned that this needs to be non-zero. So by trial and error, if I reverse and add all numbers, 92, 93, 94, 95, 96, 97, 98, 99 - only one answer is in the format CDC - i.e. 92 + 29 = 121. Rest all are not in this form so I get A=9, B = 2, C = 1 and D = 2
Now of the remaining pairs - Lets try with 8 series - Over here - I get only 1 such pair in CDC format - 83 + 38 = 121, Here A = 8, B = 3, C = 1 D = 2
Now of the remaining - Lets try with 7 series - Over here also - I get only 1 pair - 74+47 = 121. Here A = 7, B = 4, C = 1, D = 2
For 6 series - I get only 1 pair - 65 + 56 = 121. Here = A = 6, B = 5, C = 1, D = 2
Now for all the remaining series - i.e. of 5, 4, 3, 2,1 - we will not get unique values or we will get values with 0 as answer and hence we will stop here.
But in the question it is also mentioned that A, B, C and D are unique since all are denoted by separate alphabets and in our case out of the 4 possible pairs - only 3 are such pairs which are all unique. Hence the answer is Option B = 3
And we are asked the unique number of pair of A*B*C*D
This means we need to find out the total possible pairs of A, B , C and D and then the unique pairs
Now - since we are reversing the 2 digits, and the total becomes a 3 digit number, then the possible pairs can be any pairs starting from 90 to 99 except for 90 and 91 since it leads to 0 in answer and in the question it is mentioned that this needs to be non-zero. So by trial and error, if I reverse and add all numbers, 92, 93, 94, 95, 96, 97, 98, 99 - only one answer is in the format CDC - i.e. 92 + 29 = 121. Rest all are not in this form so I get A=9, B = 2, C = 1 and D = 2
Now of the remaining pairs - Lets try with 8 series - Over here - I get only 1 such pair in CDC format - 83 + 38 = 121, Here A = 8, B = 3, C = 1 D = 2
Now of the remaining - Lets try with 7 series - Over here also - I get only 1 pair - 74+47 = 121. Here A = 7, B = 4, C = 1, D = 2
For 6 series - I get only 1 pair - 65 + 56 = 121. Here = A = 6, B = 5, C = 1, D = 2
Now for all the remaining series - i.e. of 5, 4, 3, 2,1 - we will not get unique values or we will get values with 0 as answer and hence we will stop here.
But in the question it is also mentioned that A, B, C and D are unique since all are denoted by separate alphabets and in our case out of the 4 possible pairs - only 3 are such pairs which are all unique. Hence the answer is Option B = 3
11 Jul 2026, 06:15
Kudos
Bookmarks
AB+BA max can be 98+89=187
This means C=1
AB=10x+y
BA=10y+x
AB+BA=11(x+y)
121 is the only such 3 digit no.
so, D=2
Now for x+y=11
(x,y)=(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2)
As D=2, (2,9) Or (9,2) cannot be taken.
As x*y or y*x is same,
(3,8),(4,7),(5,6) satisfy
Ans B
This means C=1
AB=10x+y
BA=10y+x
AB+BA=11(x+y)
121 is the only such 3 digit no.
so, D=2
Now for x+y=11
(x,y)=(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2)
As D=2, (2,9) Or (9,2) cannot be taken.
As x*y or y*x is same,
(3,8),(4,7),(5,6) satisfy
Ans B
11 Jul 2026, 06:22
Kudos
Bookmarks
We need to understand
AB+BA=CDC
is just
10A+B + 10B+A = 101C+10D
11(A+B)=101C+10D
LHS is div by 11, so must RHS
mod 11 RHS
101 has remiander 2, and 10 has -1
2C-D is divisible by 11
Now, max distinct single digit for A and B= 9 and 8 summing to 17
A+B<=17
hence
11(17)=187>=101C+10D
possible values for C is 1 (CDC will be 101, 121, 131.. 181)
not 2, otherwise CDC become>200, our upper limit is 187.
Now 2C-D to be divisible by 11, and C=1, will have D=2
Hence
11(A+B)=121
A+B=11
remove 1,2 taken already
only pairs that satisfy are 3,8 or 4,7 or ,5,6
Hence we get products as CxD=2
AxB with the above paired up as 3x8 or 4x7 or 5x5
above will be multiplied by CxD
Which anyhow tells us there will be 3 different values
Answer B
AB+BA=CDC
is just
10A+B + 10B+A = 101C+10D
11(A+B)=101C+10D
LHS is div by 11, so must RHS
mod 11 RHS
101 has remiander 2, and 10 has -1
2C-D is divisible by 11
Now, max distinct single digit for A and B= 9 and 8 summing to 17
A+B<=17
hence
11(17)=187>=101C+10D
possible values for C is 1 (CDC will be 101, 121, 131.. 181)
not 2, otherwise CDC become>200, our upper limit is 187.
Now 2C-D to be divisible by 11, and C=1, will have D=2
Hence
11(A+B)=121
A+B=11
remove 1,2 taken already
only pairs that satisfy are 3,8 or 4,7 or ,5,6
Hence we get products as CxD=2
AxB with the above paired up as 3x8 or 4x7 or 5x5
above will be multiplied by CxD
Which anyhow tells us there will be 3 different values
Answer B
11 Jul 2026, 06:26
Kudos
Bookmarks
The C in the hundreds place of CDC can only be 1 (because two digits sum to at most 18, or 19 including the carry), so CDC can only be 121.
So A and B must add up to 11.
Possible AB values: 29 (ruled out because D=2 and all digits are distinct), 38, 47, 56, 65, 74, 83, 92 (ruled out because D=2 and all digits are distinct)
A*B*C*D = 3*8*1*2 = 48
A*B*C*D = 4*7*1*2 = 56
A*B*C*D = 5*6*1*2 = 60
A*B*C*D = 6*5*1*2 = 60
A*B*C*D = 7*4*1*2 = 56
A*B*C*D = 8*3*1*2 = 48
Different values: 48, 56, 60
IMO B
So A and B must add up to 11.
Possible AB values: 29 (ruled out because D=2 and all digits are distinct), 38, 47, 56, 65, 74, 83, 92 (ruled out because D=2 and all digits are distinct)
A*B*C*D = 3*8*1*2 = 48
A*B*C*D = 4*7*1*2 = 56
A*B*C*D = 5*6*1*2 = 60
A*B*C*D = 6*5*1*2 = 60
A*B*C*D = 7*4*1*2 = 56
A*B*C*D = 8*3*1*2 = 48
Different values: 48, 56, 60
IMO B
