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10 Jul 2026, 21:53
Kudos
Bookmarks
AB+BA=11(A+B)=CDC=101C+10D
MOD 11:
101C+10D --2C-D---0 (mod 11)
possible (C,D): (1,2), (2,4), (3,6), (4,8), (6,1), (7,3), (8,5), (9,7)
only (1,2)gives
A+B=121/11=11
since digits are distincts from 1,2
(A,B)=(3,8), (4,7) (5,6)
prod---3*8*1*2=48, 4*7*1*2=56, 5*6*1*2=60
we have 3 diff values
MOD 11:
101C+10D --2C-D---0 (mod 11)
possible (C,D): (1,2), (2,4), (3,6), (4,8), (6,1), (7,3), (8,5), (9,7)
only (1,2)gives
A+B=121/11=11
since digits are distincts from 1,2
(A,B)=(3,8), (4,7) (5,6)
prod---3*8*1*2=48, 4*7*1*2=56, 5*6*1*2=60
we have 3 diff values
10 Jul 2026, 22:55
Kudos
Bookmarks
So here what you have to do is as you can see that AB+BA Equal to CDC so what we can infer from this is that the hundredth place and the unit place are same. so just with play with certain numbers what we will do is take example of 99 if you'll add 99 +99 = 198.
Therefore this is the maximum two digit number so it is damn sure that C must be 1 and if C is 1 and obviously Adding to single digit number will give us only two digit number right like 9 + 9 is equal to 18 only so basically if we are adding unit digit B with the unit digit A then we are getting a two digit number such that their unit digit will be C (i.e. 1) it is obvious that the summation of the unit digit A + B will always give us 11 and that is our constraint.
hence we have no.s like
92
82
74
65
hence C will always be 1 and D will always be 2, and rest A& B will change making the product A*B*C*D takes 4 different values.
Ans = 4
Therefore this is the maximum two digit number so it is damn sure that C must be 1 and if C is 1 and obviously Adding to single digit number will give us only two digit number right like 9 + 9 is equal to 18 only so basically if we are adding unit digit B with the unit digit A then we are getting a two digit number such that their unit digit will be C (i.e. 1) it is obvious that the summation of the unit digit A + B will always give us 11 and that is our constraint.
hence we have no.s like
92
82
74
65
hence C will always be 1 and D will always be 2, and rest A& B will change making the product A*B*C*D takes 4 different values.
Ans = 4
10 Jul 2026, 23:15
Kudos
Bookmarks
Given,
AB = 10A +B
BA = 10B +A
So, AB + BA = 11 (A + B)
CDC is 3-digit number,
CDC = 11 (A +B)
As A, B, C, D are non-zero digits, 2 <= A +B <= 18 & A+B > 10
110 <=11(A+B) <= 198
multiple of 11 between 110 to 198 are: 110, 121, 132, 143, 154, 165, 176, 187, 198
CDC can be only 121
CDC = 11(A+B) = 121
A+B= 11
Possible pairs = (3.8), (8,3) (4,7), (7,4), (5,6), (6,5)
Therefore = A*B*C*D = A*B*1*2 = 2AB
valuse are 48, 56, 60
Answer : B (3)
AB = 10A +B
BA = 10B +A
So, AB + BA = 11 (A + B)
CDC is 3-digit number,
CDC = 11 (A +B)
As A, B, C, D are non-zero digits, 2 <= A +B <= 18 & A+B > 10
110 <=11(A+B) <= 198
multiple of 11 between 110 to 198 are: 110, 121, 132, 143, 154, 165, 176, 187, 198
CDC can be only 121
CDC = 11(A+B) = 121
A+B= 11
Possible pairs = (3.8), (8,3) (4,7), (7,4), (5,6), (6,5)
Therefore = A*B*C*D = A*B*1*2 = 2AB
valuse are 48, 56, 60
Answer : B (3)
Bunuel
