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Difficulty:
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Question Stats:
47% (01:59) correct
53%
(02:48)
wrong
based on 267
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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
_________________
+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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GMAT Club Official Solution:
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
Since the sum of two two-digit numbers is a three-digit number, the hundreds digit C must be 1. The largest possible sum is 99 + 99 = 198, so C cannot be anything greater than 1.
Now look at the units column. Since the units digit of the sum is C = 1, the sum A + B must have units digit 1. Also, since the result is three digits, A + B must be at least 10. Since A and B are digits, A + B cannot be 21 or more. Therefore:
A + B = 11
So the units column gives 1 and carries 1 to the tens column. Then the tens column is:
A + B + 1 = 12
Thus, D = 2.
So C = 1, D = 2, and A + B = 11.
Since A, B, C, and D are distinct nonzero digits, A and B cannot be 1 or 2. The possible ordered pairs for A and B are:
(3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3)
But reversed pairs give the same value of A * B * C * D, so we only need:
(3, 8), (4, 7), (5, 6)
Since C = 1 and D = 2, we get:
A * B * C * D = A * B * 1 * 2
The three possible values are:
3 * 8 * 1 * 2 = 48
4 * 7 * 1 * 2 = 56
5 * 6 * 1 * 2 = 60
So there are 3 different values.
Answer: B.
_________________
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
Since the sum of two two-digit numbers is a three-digit number, the hundreds digit C must be 1. The largest possible sum is 99 + 99 = 198, so C cannot be anything greater than 1.
Now look at the units column. Since the units digit of the sum is C = 1, the sum A + B must have units digit 1. Also, since the result is three digits, A + B must be at least 10. Since A and B are digits, A + B cannot be 21 or more. Therefore:
A + B = 11
So the units column gives 1 and carries 1 to the tens column. Then the tens column is:
A + B + 1 = 12
Thus, D = 2.
So C = 1, D = 2, and A + B = 11.
Since A, B, C, and D are distinct nonzero digits, A and B cannot be 1 or 2. The possible ordered pairs for A and B are:
(3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3)
But reversed pairs give the same value of A * B * C * D, so we only need:
(3, 8), (4, 7), (5, 6)
Since C = 1 and D = 2, we get:
A * B * C * D = A * B * 1 * 2
The three possible values are:
3 * 8 * 1 * 2 = 48
4 * 7 * 1 * 2 = 56
5 * 6 * 1 * 2 = 60
So there are 3 different values.
Answer: B.
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Here's my solution for this question
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vrinda6
Joined: 20 Jul 2020
Last visit: 05 Sep 2026
Posts: 65
Location: India
Schools: Booth '27 Kellogg '27 CBS '27 Duke '27 Haas '27 HBS '27 Ross '27 Stern '27 Stanford '27 Wharton '27 Anderson '27
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Schools: Booth '27 Kellogg '27 CBS '27 Duke '27 Haas '27 HBS '27 Ross '27 Stern '27 Stanford '27 Wharton '27 Anderson '27
Posts: 65
Start with the equation.
AB=10A+ B, BA= 10B+A
ALSO
CDC= 100C + 10D + C= 102C+10D
Solving for both, we have:
11(A+B)= 102C+ 10D
And considering A and B are ditinct digits
A+B<=9+8= 17
thus
11(A+B)<=11X17=187 ( if c>=2 then cdc>=202 which is not possible)
So we get c=1
Substituing C in 11(A+B)=102+10D we get D=3
Now the only thing remaining is finding the possible pairs.
sum needs to be 12, and digits need to be distinct which cannot be 1 or 3
A,B= (4,8) OR (8,4)
A,B= (5,7) OR (7,5)
Finding the product:
4x8x1x3= 96
5x7x1x3=105
hence correct answer a=2
AB=10A+ B, BA= 10B+A
ALSO
CDC= 100C + 10D + C= 102C+10D
Solving for both, we have:
11(A+B)= 102C+ 10D
And considering A and B are ditinct digits
A+B<=9+8= 17
thus
11(A+B)<=11X17=187 ( if c>=2 then cdc>=202 which is not possible)
So we get c=1
Substituing C in 11(A+B)=102+10D we get D=3
Now the only thing remaining is finding the possible pairs.
sum needs to be 12, and digits need to be distinct which cannot be 1 or 3
A,B= (4,8) OR (8,4)
A,B= (5,7) OR (7,5)
Finding the product:
4x8x1x3= 96
5x7x1x3=105
hence correct answer a=2
ankushsambare
Joined: 13 Jun 2022
Last visit: 13 Sep 2026
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Schools: ISB '27 Wharton '26 INSEAD Aug '26
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Given
AB
+ BA
CDC
write in term of unit digits
10a+b +10b+a = 11(a+b)
now, Unit digit
a+b = c+10k
Tens
a+b+k = 10c+d
since sum is 3 digit no, the carry into the hundereds is 1,
so
c=1
thus
a+b = 1+10k
since a & b are non zero digits, a + b cannot equal 1 so k =1
therefore a+b =11
substitute into tens eq
11+1 = 10+d
d=2
thus, c=1 & d=2
the distinct digit pairs with sum 11 are
(2,9),(3,8),(4,7),(5,6)
Reject (2,9) because 2 is already used as d
remaining pairs
reversing a n b give the same products
so only 3 different products
Option:B
AB
+ BA
CDC
write in term of unit digits
10a+b +10b+a = 11(a+b)
now, Unit digit
a+b = c+10k
Tens
a+b+k = 10c+d
since sum is 3 digit no, the carry into the hundereds is 1,
so
c=1
thus
a+b = 1+10k
since a & b are non zero digits, a + b cannot equal 1 so k =1
therefore a+b =11
substitute into tens eq
11+1 = 10+d
d=2
thus, c=1 & d=2
the distinct digit pairs with sum 11 are
(2,9),(3,8),(4,7),(5,6)
Reject (2,9) because 2 is already used as d
remaining pairs
reversing a n b give the same products
so only 3 different products
Option:B
Since its a 2 digit addition, the maximum value for c can be 1
Since even 99 + 99 = 198
Thus the addition of B + A must end in 1
Therefore B + A has to be 11
Possible pairs = (2,9) (3,8) (4,7) (5,6)
Now for D the 1 from 11 will be carried over
Therefore the equation will be
1 + A + B = D (Substitute earlier equation)
1 + 11 = D
12 = D
Since A, B, C, D are distinct values we can eliminate the (2,9)
Thus, 3 values are possible (B)
Since even 99 + 99 = 198
Thus the addition of B + A must end in 1
Therefore B + A has to be 11
Possible pairs = (2,9) (3,8) (4,7) (5,6)
Now for D the 1 from 11 will be carried over
Therefore the equation will be
1 + A + B = D (Substitute earlier equation)
1 + 11 = D
12 = D
Since A, B, C, D are distinct values we can eliminate the (2,9)
Thus, 3 values are possible (B)
Yesterday's PS questions were so hard, it is nice to have a bit of a break from that.
I selected Choice B.
Rewrite the given expression and examine it by inspection to see what must be true about the given digits - we know that they must be distinct nonzero integers which narrows it down quite a bit.
We know that B + A = C, but also that A + B + 1 = CD, because if A + B < 10, then there would not be an additional digit D. So we can safely say that A + B > 10.
First I tried A = 8 and B = 9, which is 89+98 = 187. I quickly realized that in order for C to appear twice, C would need to be equal to 1, because we are only able to add 1 to the tens digits and the sum of the tens digits plus one can actually never exceed 1.
Therefore, try numbers which are greater than 10 and end in 1, i.e., 11.
We find that pairs of digits for A & B which add to 11 work, namely, (2,9), (3,8), (4,7) and (5,6). Due to the question asking for the number of unique products, the order of these does not matter, due to the commutative property of multiplication. Note that 2,9 is not valid because it would yield a value of 2 for D, which would violate the rule that all digits must be distinct. Take care to ensure all cases are valid before counting them. Therefore, there are 3 different values and the answer is B.
I selected Choice B.
Rewrite the given expression and examine it by inspection to see what must be true about the given digits - we know that they must be distinct nonzero integers which narrows it down quite a bit.
We know that B + A = C, but also that A + B + 1 = CD, because if A + B < 10, then there would not be an additional digit D. So we can safely say that A + B > 10.
First I tried A = 8 and B = 9, which is 89+98 = 187. I quickly realized that in order for C to appear twice, C would need to be equal to 1, because we are only able to add 1 to the tens digits and the sum of the tens digits plus one can actually never exceed 1.
Therefore, try numbers which are greater than 10 and end in 1, i.e., 11.
We find that pairs of digits for A & B which add to 11 work, namely, (2,9), (3,8), (4,7) and (5,6). Due to the question asking for the number of unique products, the order of these does not matter, due to the commutative property of multiplication. Note that 2,9 is not valid because it would yield a value of 2 for D, which would violate the rule that all digits must be distinct. Take care to ensure all cases are valid before counting them. Therefore, there are 3 different values and the answer is B.
Archit3110
Major Poster
Joined: 18 Aug 2017
Last visit: 13 Sep 2026
Posts: 8,817
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73 
GMAT Focus 2: 645 Q83 V82 DI81 
GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,817
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
from given addition value of
10a+b+10b+b ; 11( a+b)
where sum value is CDC ; possible when a+b is 11
11*11 = 121
2 digit numbers whose sum is 11
a,b possible 38,47,56 & its
ba 83,74,65
different '3' values will be possible when a*b*c*d
OPTION B ; 3 is correct
+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
from given addition value of
10a+b+10b+b ; 11( a+b)
where sum value is CDC ; possible when a+b is 11
11*11 = 121
2 digit numbers whose sum is 11
a,b possible 38,47,56 & its
ba 83,74,65
different '3' values will be possible when a*b*c*d
OPTION B ; 3 is correct
JollyScale
Joined: 04 Jul 2026
Last visit: 13 Sep 2026
Posts: 55
Location: India
Schools: HBS '27 CBS '27 Stanford '27
GPA: 6.87/7
WE:Marketing (Finance: Diversified Financial Services)
Products:
Schools: HBS '27 CBS '27 Stanford '27
Posts: 55
Since AB and BA are two digit it will be can be from 10 to 99
But also consider that their addition should result in a 3 digit nos.
which can be from 100 to 198
since it is CDC it can be 1D1
Now AB + BA = (10A + B) + (10B + A) = 11(A+B)
meaning 1D1 must be divisible by 11 which is 121
as 11(A+b) = 121
A+B = 11
pssobile ans (3 +8) (4+7) (5+6)
3*8*1*2
4*7*1*2
5*6*1*2
We get 48, 56 & 60
3 values
Ans is B
But also consider that their addition should result in a 3 digit nos.
which can be from 100 to 198
since it is CDC it can be 1D1
Now AB + BA = (10A + B) + (10B + A) = 11(A+B)
meaning 1D1 must be divisible by 11 which is 121
as 11(A+b) = 121
A+B = 11
pssobile ans (3 +8) (4+7) (5+6)
3*8*1*2
4*7*1*2
5*6*1*2
We get 48, 56 & 60
3 values
Ans is B
Bunuel
Kinshook
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Schools: HBS '28 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
A B
+ B A
_____
C D C
10A + B + 10B + A = 11 (A+B) = 100C + 10D + C
Since A, B, C & D are non-zero digits.
11 (A + B) > 111
A + B > 111/11 = 10
Case 1: A + B = 11
1a : A = 2; B = 9; AB + BA = 121; C = 1; D = 2; A=D=2 Not feasible since they are all distinct
1b : A = 3; B = 8; AB + BA = 121; C = 1; D = 2; Feasible; A*B*C*D = 3*8*1*2 = 48
1c : A = 4; B = 7; AB + BA = 121; C = 1; D = 2; Feasible; A*B*C*D = 4*7*1*2 = 56
1d: A = 5; B = 6; AB + BA = 121; C = 1; D = 2; Feasible; A*B*C*D = 5*6*1*2 = 60
For rest of the cases A & B are swapped but the product A*B*C*D remained unchanged.
Case 2: A + B = 12; 11 (A+B) = 132; Not feasible since 11 (A+B) is not of the form CDC
Case 3: A + B = 13; 11 (A+B) = 143; Not feasible since 11 (A+B) is not of the form CDC
Similarly 14*11 = 154; 15*11= 165; 16*11=176; 17*11 = 187; Not feasible
A*B*C*D can take 3 different values e.g. 48, 56 & 60.
IMO B
+ B A
_____
C D C
10A + B + 10B + A = 11 (A+B) = 100C + 10D + C
Since A, B, C & D are non-zero digits.
11 (A + B) > 111
A + B > 111/11 = 10
Case 1: A + B = 11
1a : A = 2; B = 9; AB + BA = 121; C = 1; D = 2; A=D=2 Not feasible since they are all distinct
1b : A = 3; B = 8; AB + BA = 121; C = 1; D = 2; Feasible; A*B*C*D = 3*8*1*2 = 48
1c : A = 4; B = 7; AB + BA = 121; C = 1; D = 2; Feasible; A*B*C*D = 4*7*1*2 = 56
1d: A = 5; B = 6; AB + BA = 121; C = 1; D = 2; Feasible; A*B*C*D = 5*6*1*2 = 60
For rest of the cases A & B are swapped but the product A*B*C*D remained unchanged.
Case 2: A + B = 12; 11 (A+B) = 132; Not feasible since 11 (A+B) is not of the form CDC
Case 3: A + B = 13; 11 (A+B) = 143; Not feasible since 11 (A+B) is not of the form CDC
Similarly 14*11 = 154; 15*11= 165; 16*11=176; 17*11 = 187; Not feasible
A*B*C*D can take 3 different values e.g. 48, 56 & 60.
IMO B
Given that AB + BA = CDC, this means C must be 1 because sum of any two digit numbers can not exceed 199.
Now we know C is also in ones digit means sum of A & B must be 11, so it C can be at both ones and hundreth digit.
Thus A & B can (5,6) (4,7) ,(3,8),(2,9),(9,2),(8,3),(7,4),(6,5)
And if we add these we always get the sum be 121. thus D must be 2 only.
Now we know product 5*6*1*2=6*5*1*2
thus we have only 4 values for A*B*C*D
Ans = 4.
Now we know C is also in ones digit means sum of A & B must be 11, so it C can be at both ones and hundreth digit.
Thus A & B can (5,6) (4,7) ,(3,8),(2,9),(9,2),(8,3),(7,4),(6,5)
And if we add these we always get the sum be 121. thus D must be 2 only.
Now we know product 5*6*1*2=6*5*1*2
thus we have only 4 values for A*B*C*D
Ans = 4.
i think this is 3?
given AB +BA = CDC (all are distint nonzero)
so 10A + B + 10B + A = 100 C + 10D + C
11A + 11B = 101C + 10D - Eq1
AB max can be 99 - so sum = 99 + 99 = 198, considering this, C can never be greater than 2 or 2 itself
so C = 1
sub in eq 1
11 (A +B) = 101 + 10D
101 + 10D should be divisible by 11, as A+B cannot be decimal or Fraction
so D can only be 2
121/11 = 11
A+B = 11
so possible pairs are - (2,9) (3,8) (4,7) (5,6) (6,5) (7,4) (8,3) (9,2)
but it said that all are distinct as D = 2, (2,9) and (9,2) these two pairs are not possible
similarly, the question asked for how many A*B*C*D = 2 *A *B - sub the other pairs, we get 3 distinct values
as 3,8 or 8,3 will yield same results, so consider as one and apply for the other 4 pairs too
given AB +BA = CDC (all are distint nonzero)
so 10A + B + 10B + A = 100 C + 10D + C
11A + 11B = 101C + 10D - Eq1
AB max can be 99 - so sum = 99 + 99 = 198, considering this, C can never be greater than 2 or 2 itself
so C = 1
sub in eq 1
11 (A +B) = 101 + 10D
101 + 10D should be divisible by 11, as A+B cannot be decimal or Fraction
so D can only be 2
121/11 = 11
A+B = 11
so possible pairs are - (2,9) (3,8) (4,7) (5,6) (6,5) (7,4) (8,3) (9,2)
but it said that all are distinct as D = 2, (2,9) and (9,2) these two pairs are not possible
similarly, the question asked for how many A*B*C*D = 2 *A *B - sub the other pairs, we get 3 distinct values
as 3,8 or 8,3 will yield same results, so consider as one and apply for the other 4 pairs too
Posts: 60
i am going with option b.
c and d are fixed, c=1, d=2 as 121 is the only possible sum
a and b must add to 11, they cannot be 1 or 2 as digits must be different
3 pairs - 3/8, 4/7, 5/6 yielding products 48, 56 and 60
a incorrect as it forgets the pair 5/6
c incorrect it includes the pair 2/9 which is wrong as d is 2 already
d and e incorrect if you count reverse pair 3/8, 8/3 as separate products as it is not.
c and d are fixed, c=1, d=2 as 121 is the only possible sum
a and b must add to 11, they cannot be 1 or 2 as digits must be different
3 pairs - 3/8, 4/7, 5/6 yielding products 48, 56 and 60
a incorrect as it forgets the pair 5/6
c incorrect it includes the pair 2/9 which is wrong as d is 2 already
d and e incorrect if you count reverse pair 3/8, 8/3 as separate products as it is not.
since AB +BA =CDC, C has to be 1 as both left and right C value will be same. D=2 because of carry forward.
Different values of 11 are 2&9, 3&8, 4&7, 5&6.
so different product of A,B, C, D are 4.
Answer(C)
Different values of 11 are 2&9, 3&8, 4&7, 5&6.
so different product of A,B, C, D are 4.
Answer(C)
Posts: 16
AB + BA = CDC
That means, the ones place and the hundred place should have the same number - C
And we know the max number will be less than 200 (99 + 99) = 198
So we have to make the sum such that the ones and the third place is 1.
Now, we know 5 + 6 = 11, 7 + 4 = 11, 8 + 3 = 11, 9 + 2 = 11
once we substitute - (56,65) - 121(CDC - follow), (74, 47 - follow) - 121(CDC - follow), (83, 38) - 121(CDC - follow), (92, 29) - 121(CDC -but cannot follow since D = B. here and in ques it's mentioned all numbers different)
so, the number can only. be - (5,6), (7,4), (8,3)
Also, it can be interchanged to get all the combinations
But in question it's asking A* B* C* D - thus, interchanging A, B will give the same answer
Hence, only 3 combinations are possible.
Answer, B is correct
That means, the ones place and the hundred place should have the same number - C
And we know the max number will be less than 200 (99 + 99) = 198
So we have to make the sum such that the ones and the third place is 1.
Now, we know 5 + 6 = 11, 7 + 4 = 11, 8 + 3 = 11, 9 + 2 = 11
once we substitute - (56,65) - 121(CDC - follow), (74, 47 - follow) - 121(CDC - follow), (83, 38) - 121(CDC - follow), (92, 29) - 121(CDC -but cannot follow since D = B. here and in ques it's mentioned all numbers different)
so, the number can only. be - (5,6), (7,4), (8,3)
Also, it can be interchanged to get all the combinations
But in question it's asking A* B* C* D - thus, interchanging A, B will give the same answer
Hence, only 3 combinations are possible.
Answer, B is correct
Answer: B) 3
Posts: 112
AB
+BA
-------
CDC
Look at ones column
B + A + C + 10(Carry over)
Look at 10s column
A + B + (Carry over) = D + 10(Carry over)
Maximun A and B can be is 9 and 8 with 1 as carry over = 18
Therefor C must be 1
A + B = C Since C =1, A and B have to be greater than or equal to 2, therefor ones column also carries over a 1
A + B = 11
so either 3,8 4,7 or 5,6
Tens column must be 11 + 1 = D + 10
D =2
Therefore A,B, C, D
is either
3,8,1,2 4,7,1,2 or 5,6,1,2
reversing A and B does not change the product
Answer B 3
+BA
-------
CDC
Look at ones column
B + A + C + 10(Carry over)
Look at 10s column
A + B + (Carry over) = D + 10(Carry over)
Maximun A and B can be is 9 and 8 with 1 as carry over = 18
Therefor C must be 1
A + B = C Since C =1, A and B have to be greater than or equal to 2, therefor ones column also carries over a 1
A + B = 11
so either 3,8 4,7 or 5,6
Tens column must be 11 + 1 = D + 10
D =2
Therefore A,B, C, D
is either
3,8,1,2 4,7,1,2 or 5,6,1,2
reversing A and B does not change the product
Answer B 3
AB = 10A+B; BA = 10B +A; CDC = 100C+10D+C;
11(A+B) = 101C + 10D....i
now, as A and B are single non zero distinct digits - max value of them can be 9 and 8 making lhs = 187. therefore C can only be 1. so,
11(A+B) = 101 + 10D; now lhs is a multiple of 11 so should be the rhs;
looking for relevant values:
D = 2 is the only solution for which rhs is multiple of 11. so D = 2;
so the rhs number is 121. therefore solving i eq after adding values of C and D
A+B = 11. we don't have any other constraints so possible values of A and B are - (3,8),(4,7),(5,6) and the mirror set(not 2 or 1 - distinct). we need product so that is not going to effect our final solution.
Therefore, A*B*C*D can have 3 different values(as C and D have only have 1)
Therefore, Answer = B
11(A+B) = 101C + 10D....i
now, as A and B are single non zero distinct digits - max value of them can be 9 and 8 making lhs = 187. therefore C can only be 1. so,
11(A+B) = 101 + 10D; now lhs is a multiple of 11 so should be the rhs;
looking for relevant values:
D = 2 is the only solution for which rhs is multiple of 11. so D = 2;
so the rhs number is 121. therefore solving i eq after adding values of C and D
A+B = 11. we don't have any other constraints so possible values of A and B are - (3,8),(4,7),(5,6) and the mirror set(not 2 or 1 - distinct). we need product so that is not going to effect our final solution.
Therefore, A*B*C*D can have 3 different values(as C and D have only have 1)
Therefore, Answer = B
Posts: 73
IMO : 3
AB + BA = CDC
also note that AB and BA are two digit numbers
so we can write them as AB ====> (10A +B ) and similarly BA =====> 10B+A
now adding them will give 11(A+B) which as per the question is equal to CDC
bow note here that CDC has to be a multiple of 11 and also that their unit and hundred digit will be same,
C =A+B
but will give some carry forward coz this thime B+A + carry = D
so now using a multiple of 11 which is 1st palindrom --- >121
here we see that A+B =11 , also that C =1 , D =2
now possible cases :
A B possible
9 2 no as all are distinct and D =2
8 3 yes
7 4 yes
6 5 yes
next they will repeat since we need multiplication
hence 3 possible values
AB + BA = CDC
also note that AB and BA are two digit numbers
so we can write them as AB ====> (10A +B ) and similarly BA =====> 10B+A
now adding them will give 11(A+B) which as per the question is equal to CDC
bow note here that CDC has to be a multiple of 11 and also that their unit and hundred digit will be same,
C =A+B
but will give some carry forward coz this thime B+A + carry = D
so now using a multiple of 11 which is 1st palindrom --- >121
here we see that A+B =11 , also that C =1 , D =2
now possible cases :
A B possible
9 2 no as all are distinct and D =2
8 3 yes
7 4 yes
6 5 yes
next they will repeat since we need multiplication
hence 3 possible values
10 A + B + 10B + A = 100C + 10D + C
We know C has to be 1 (even if the 2 digits nos were 98 + 89, total cannot exceed 199)
11A + 11B = 101 + 10 D
11 (A+ B) = 101 + 10 D
D has to be such a no that it makes whole right side multiple of 11
D can be 2 and no other number, C = 1, D = 2; A and B total 11, they can be 5,6 / 4, 7 / 3, 8 / 2, 9 not possible as D is 2. So 3.
We know C has to be 1 (even if the 2 digits nos were 98 + 89, total cannot exceed 199)
11A + 11B = 101 + 10 D
11 (A+ B) = 101 + 10 D
D has to be such a no that it makes whole right side multiple of 11
D can be 2 and no other number, C = 1, D = 2; A and B total 11, they can be 5,6 / 4, 7 / 3, 8 / 2, 9 not possible as D is 2. So 3.
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