AB +BA ____ CDC In the correctly worked addition problem shown, AB and BA are two-digit positive integers, CDC is a three-digit integer, and A, B, C, and D are distinct nonzero digits. How many different values can A * B * C * D take? A. 2 B. 3 C. 4 D. 6 E. 8 ⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban. M46-39

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GMAT Club World Cup 2026 (Day 5): In the correctly worked addition

1 2 3 4 5

Saurav1408

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10 Jul 2026, 09:04

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In this question, we need to note that - A, B, C, D are DISTINCT and NON ZERO and AB and BA are both two digit positive integers and CDC is a three digit integer.

if we start with
AB = 21, then we would quickly realize that this 21 + 12 would not give three digit integer. In order to get 3 digit integer the sum of the two unit digits B and A shall be greater than 10 (as C cannot be zero too)

Make the sum 11 of the two unit digits and write the pairs that would give unit digit sum = 11
So starting with
Case 1: AB = 56, 56 + 65 = 121; all different and non zero
Case 2: AB = 47, 47 + 74 = 121; all different and non zero
Case 3: AB = 38, 38 + 83 = 121; all different and non zero
Case 4: AB = 29, 29 + 92 = 121; all are not different here A and D are both equal to 2 so this case eliminated

We can get CDC type number if we make the sum 131 but for 131 we need to carry forward 2 from the unit digits addition to the ten's digit addition and it is not possible as maximum sum we could get from 2 unit digits is 17 = 9 + 8

So only 3 cases are possible and hence answer is B. 3

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10 Jul 2026, 09:05

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Given
AB
+BA
------
CDC

the maximum value of B+A is 9+8 = 17. If that case is considered, then it is clear that value of C at max can be 1 only. Because the left most C in the answer above is obtained by the carry from the sum of left most A+B+ carry over from right most B+A.

Thus if C=1, then B+A =11 ( For the right most B+A to give a value of 1 in the answer ( C) , their sum has to be 11).

Hence we just need to find of unordered pairs of A,B whose sum is 11.
hence (2,9), (3,8), (4,7), (5,6) are the only pairs that satisfy this. Hence the ans is 4

Bunuel

AB
+BA
____
CDC

In the correctly worked addition problem shown, AB and BA are two-digit positive integers, CDC is a three-digit integer, and A, B, C, and D are distinct nonzero digits. How many different values can A * B * C * D take?

A. 2
B. 3
C. 4
D. 6
E. 8

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10 Jul 2026, 09:08

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This is how I solved it.
The answers could include other options but then all A, B, C and D won't be distinct
Hence, the answer is 3

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10 Jul 2026, 09:12

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AB+BA = CDC

The only way C can be is 1, so AB + BA = 1D1

To end up with 1 as the last digit, there are 4 pairs of digits to plug in.
1. 2 and 9 -> 29+92 = 121, which is not possible since D and (A or B) are different digits.
2. 4 and 7 -> 47+74 = 121, yes!
3. 6 and 5 -> 65+56 = 121, yes!
4. 8 and 3 -> 83+38 = 121, yes!

So, we know that
A and B: 4 possible digits
C and D: 1 possible digit

To find the value A*B*C*D can take. C and D don't add anything, so we can focus on A and B.

3*7 = 7*3 -> one
6*5 = 5*6 -> two
8*3 = 3*8 -> three

So, choice B

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i think that might be
some sort of reversing like 19 and 91
so that if repacing tham so 2 different possible answer

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10 Jul 2026, 09:27

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A+B needs to = 11 for the result to come out as 121 or CDC
the possible pairings are 2,9 or 3,8 or 4,7 or 5,6
BUT the statement says A,B,C,D are DISTINCT, hence, the pair 2,9 won't be possible since D already = 2

so there are total 3 possible pairs and hence 3 values of A x B x C x D

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10 Jul 2026, 09:34

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1. Set up the equation algebraically:
$AB + BA = CDC$
$(10A + B) + (10B + A) = CDC$
$11(A + B) = CDC$

2. Deduce C and D:
Because $A$ and $B$ are distinct single digits, their maximum sum is 17 (9 + 8).
The maximum possible value for $CDC$ is $11 \times 17 = 187$.
Therefore, C = 1.
The only multiple of 11 in the $1\_1$ format is $121$.
Therefore, D = 2 (and $A + B = 11$).

3. Find valid pairs for A and B:
We need pairs that sum to 11. Since $A$, $B$, $C$, and $D$ must be distinct, we cannot use 1 or 2.

$9 + 2$ (Invalid: 2 is already used for D)
$8 + 3$ (Valid)
$7 + 4$ (Valid)
$6 + 5$ (Valid)

4. Calculate the possible products ($A \times B \times C \times D$):

$8 \times 3 \times 1 \times 2 =$ 48
$7 \times 4 \times 1 \times 2 =$ 56
$6 \times 5 \times 1 \times 2 =$ 60

There are exactly 3 possible values for the product.

Correct Answer: B

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10 Jul 2026, 09:38

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AB+BA=CDC C can only be 1... take combos of no.s tht give 1 at end, 5,6; 7,4; 9,2we have below
so answer si 3 (B)

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AB+BA=CDC
The maximum possible values of a 2 digit number is 99 (99+99=198 is the highest possible value of sum)
CDC is a 3 digit number so C must be 1.
The sum of units digits A+B must end in 1. And only possible sums of 2 digits that end in 1 is 11. Then, A+B must be 11.
A and B can be (9,2), (8,3), (7,4), (5,6).
Put them in AB+BA=1D1 and check
92+29= 121. (A,B,C,D are different digits.) Not possible
83+38= 121
74+47=121
65+56= 121.
We need to find A*B*C*D
1. 1*2*3*8= 48
2. 1*2*4*7= 56
3. 1*2*5*6= 60
3 different values are possible

B

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C has to be 1 irrespective of any value of A or B.
As A+B cannot be equal to 1, A+B has to add up to 11.
If A+B=11, D will always be 2.
Given the above, C and D are always 1 and 2 respectively and neither A nor B can be 1 or 2.

Since different values of A*B*C*D are in question, the order of (A,B) does not matter, i.e. (9,2) and (2,9) will give us the same value.

Possible values of (A,B) which provide a distinct value: (3,8),(4,7),(5,6)
Number of distinct values = 3

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If you think about this carefully, you will realize that the only way to get a hundreds digit when adding 2 two-digit numbers is from the carry out of the tens column.
The carry can only be 1 (since 9+9+1 = 19)
Therefore, C = 1.

The units digit is also C, i.e., 1. To get a units digit = 1, A+B = 11 (because digits are non-zero, A+B = 1 is impossible)
So, A+B = 11

The tens column adds to: A+B+Carry = 11+1 = 12
So, the tens digit is D = 2
Hence, now, C = 1, D = 2 and A+B = 11.

Now, find all possible pairs: Distinct non-zero digits summing to 11:
(3,8), (4,7) and (5,6)
(2,9) is not valid as D = 2, and all digits must be distinct

Calculating the products to find the unique number of possible products => 48, 56 and 60

Thus, there are 3 different products possible.

Hence, the correct answer is Option B - 3.

Hope this helps!

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Answer is (B) 3.

We are given that,
AB
+ BA
----------
CDC

By rules of addition, there are two possibilities here,
Either,
B+A = C (which is the sum of A+B does not exceed 10 and hence 1 is not carried over during the sum of tens places)
OR
B+A = 10 + C (which is sum of A+B is greater than 10 and hence 1 will be carried over during sum of tens places)

Let's try and explore both options:
Option 1: B+A = C
This would mean that 1 is not carried over during the sum of the tens places
Hence, sum of tens places would give us the number CD (10C + D)
A+B = 10C +D

Now if we equate both eqns, we get
C = 10C + D
or
-9C = D
since C and D are digits in a number they cannot be negative.
Hence our option 1 is incorrect and cannot be the case.

Option 2: B+A = 10 + C
This would mean that 1 is carried over during the sum of tens place.
Thus,
A+B+1 = 10C + D
Substituting the value of A+B, we get
10 + C + 1 = 10C + D
or
11 = 9C + D

Since all A, B, C and D are nonzero digits, the only possible combination of C and D that satisfies the equation is C = 1 and D = 2
If we substitute the value of C in A+B = 10 + C,
we get A+B = 11.

we are also given that A, B, C and D are distinct. Hence the only possible combinations that satisfy A+B = 11 are,
(3,8), (4,7), (5,6) ___ (C and D are already 1 and 2 hence we ignore (2,9) )
Also, order does not matter since ultimately we need the product of A*B*C*D.
Since there are 3 pairs we can have 3 distinct products.

Hence (B) is the correct answer choice

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Another new topic for me, so I used a process that relied on logic.
Happy to see a faster method!

Since we need 2 digit number adding to three digit numbers, and A+B can NOT give zero in units place, the following combos are eliminated-
(A,B) can not be (1,9) (2,8) (3,7) (4,6)

I started with 51 and worked my way, 51+15 is under 100. This goes on until we reach 55, but A and be are UNIQUE integers. So we test 56 onwards.
If (AB)=56
56+65=121. (VALID)
57+75=132 (INVALID since we need CDC)
58+85=143 (INVALID)
59+95=154 (INVALID)
60+06= INVALID since all terms are non-zero.
(We see increase of 11, which makes sense since we are increasing first term by '1' and by default second term will increase by '10')

Similarly for 60-70 range, we start with 64
64+46=110 (INVALID)
65+56=121 (VALID)
66+66=132 (INVALID since each integer is unique)
67+76=143
We note similar progression, 121 and then and increase of 11, and invalid results like above. This establishes a clear pattern.

By the time we reach 70-80 range, we notice another pattern. For 50-60 sum went beyond 100 starting 55, for 60-70 it went beyond 100 starting 64. So we test 73 directly and see that it is indeed true as 72+27=99
73+37=110 (INVALID)
74+47=121 (VALID)
75+57=132 (INVALID)
(Now we see an increment of 11, and it will follow a similar progression as above and give invalid outputs that don't match the CDC pattern)

For 80-90, we test from 82
82+28=110 (INVALID)
83+38=121 (VALID)
84+48= 132 (INVALID)
(Moving forward will continue giving invalid outputs as we have seen above)

For 90-99, we test from 91
91+19=110 (INVALID)
92+29=121
(We know that beyond this it will go invalid)

So far the valid options are-
56+65=121
65+56=121
74+47=121
83+38=121
92+29=121
However, since A,B,C,D are unique integers, 92+29=121 becomes invalid as it would result in A=D (both will be 2)
And since 56

Therefore A can be: (5,6),(6,5),(7,4),(8,3)
B can be: 6,5,4,3
(C,D) can be ONLY (1,2)

C*D will always be 2.

(A*B) can be-
(5*6)=30
(6*5)=30
(7*4)=28
(8*3)=24
Since, first two are the same, we can count it as one.

Possible values of A*B*C*D-
30*2=60
28*2=56
24*2=48

Total possible values-3

Ans- (B)

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Given: AB + BA = CDC and A, B, C and D each have distinct positive, non- zero integer values.

Since the value obtained upon adding AB and BA is CDC i.e., has 3 digits, the sum of the digits A and B has to be >=10 otherwise the result of the addition would only have 2 digits.
But since neither of the digits in the whole operation is 0, A + B > 10.

It's also important to note that it's not possible for C to be 2 since we only add two digits A and B and the addition of 2 single-digit numbers is always <= 18 (or 9 + 9). The other case where the units digit could be 2 is if A + B = 12 but that's not going to give us a valid solution since the addition of AB + BA in the case where the units digit is 2 will be 132 which is not in the form of CDC. We can eliminate other digits (except 1) with the same logic.

Therefore, the only way to add two 2-digit numbers where the units and hundredths positions of the resultant value are the same is if these two places have the value 1. Hence, A + B = 11.

Now, the ways to add 2 single-digit numbers where the result will be 11 is [9 and 2], [8 and 3], [7 and 4] and [5 and 6].
However, if A is 9 and B is 2 (or vice-versa) D would also be 2 which is not permitted according to the constraint mentioned in the question where A, B, C and D all hold distinct values. Hence, [9 and 2] is not a valid solution.
Therefore, the only distinct valid solutions to the problem are [8 and 3], [7 and 4] and [5 and 6].
Hence, A * B * C * D can have 3 distinct values where C = 1 and D = 2.
Final answer: B. 3

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two 2 digit numbers + to 3 digit number is carry over 100 digit must be 1. so C=1
B+A last number C
A and B are non zero digits, the sum must be 11
finding D
A+B+carry of 1=11+1=12
so the tens number must be 2, making D=2
A+B=11
C=1 and D=2, A and B cannot be 1 or 2
Remaining number pairs that sum to 11 are (3,8), (4,7), (5,6)
3*8*1*2=48
4*7*1*2=56
5*6*1*2=60
ANS:B 3

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Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

we can write the two numbers as, 10A+B, 10B+A
their sum is : 11(A+B) = CDC
given that A,B are distinct non zero digits, we can say the maximum value of A,B = 9,8
So, A+B = 17
so, 11(A+B)= 11(17) =187
So, max value of CDC = 187
So, C = 1 ( CDC is a 3 digit number)
then 1D1 and multiple of 11 is only possible when D =2
so, CDC = 121
A+B=11
as A,B,C,D are all distinct non zero - A,B cannot be 1,2
then, possible values of A,B that satisfy, A+B = 11
3+8
4+7
5+6
then, different values of A*B*C*D =
(i) 3*8*1*2 = 48
(ii) 4*7*1*2= 56
(iii) 5*6*1*2=60
therefore, the no of possible values of A*B*C*D is 3
B

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10 Jul 2026, 10:41

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AB=10A+B
BA=10B+A
AB+BA=11(A+B)

CDC=100C+10D+C=101C+D
101C+10D=11(A+B)

OHH LEFT SIDE SEEMS DIVISIBLE BY 11.. SO RIGHT SHOULD ALSO BE

DIVISIBILITY CONDITION OF 11, 2C-D SHOULD BE DIVISIBLE BY 11 OR 0

IF 0 THEN 2C=D THAT MEANS 121C =11(A+B) THAT MEANS 11C=A+B
ALL DISTINCT SO ONLY POSSIBLE IF C=1 SINCE C=2 WILL MAKE A .B DOUBLE DIGIT INTERGER BUT THEY ARE SINGLE DIGIT INTEGER

S0 A+B=11 , C=1 AND D=2 CDC=121 (A, B) CAN BE (3,8), (4,7), (5,6) AND EXCHANGEABLY... (2,9) PAIR WILL MAKE D NON-DISTINCT.

SO THREE DIFFERENT PRODUCTS

IF 11 THEN, 2C-D=11, D =11-2C, THIS CASE DO NOT WORK

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AB = 10A + B and BA = 10B + A

AB + BA = 11A + 11B = 11 (A+B)

So, CDC is a three digit positive integer that must be divisble by 11

Now, the largest sum of AB + BA can be 98 + 89 = 187

So, from 110 to 187, only 121 can be CDC

So, A and B can not be 1 and 2
We start with A = 3, for c = 1, the only possible value of B is 8, so 38 + 83 = 121
A = 4, B = 7 (to make c = 1), so 47 + 74= 121
A = 5, B = 6
No other value is possible from here on

So, 3 distinct product values of A*B*C*D are possible

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I can write AB as 10A+B and BA as 10B+A. Their sum would then be 10A+B+10B+A = 11A+11B = 11(A+B). So the sum of these two integers will be a multiple of 11 (which in turn means CDC will be a multiple of 11)

The maximum value that A and B can take is 9 and 8 respectively (or I can assign 8 to A and 9 to B, doesn't really matter). So maximum AB+BA = 11(A+B) = 11(9+8) = 187

The minimum value 11(A+B) can assume is 110 (which will be 11*10)
So between 110 and 187, we have following multiples of 11 = 110, 121, 132, 143, 154, 165, 176, 187. Only 121 is in the form CDC. Hence 11(A+B) = 121 or A+B=11, C = 1 and D = 2

For sum of A & B to be 11, A & B can take following values = 9&2, 8&3, 7&4 and 6&5. Since A,B,C and D are all distinct digits, 9&2 is not a valid combination (because D=2). Hence there are 3 possible combinations of A*B*C*D (viz, 8*3*1*2, 7*4*1*2 and 6*5*1*2)

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at units digit, B+A = C. we donot know if there is a carry or not.
at tens digit, A+B = CD hence there was a carry earlier.

C cannot be 0 since three digit number CDC cannot begin with 0 and question says all are non-zero.
A+B results in two digit number: least can be 1+9 = 10 and max can be 8+9 = 17. each of these values has tens digit of 1 hence C = 1.
so A+B = B+A is a two digit number which has both unit digit and tens digit of 1 which means D = 2.
since all digits are distinct, A & B can take following values which add up to 11: 3,8 or 4,7 or 5,6

A*B*C*D = 3*8*1*2=48 or 4*7*1*2=56 or 5*6*1*2=60
there are 3 possible values. B