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10 Jul 2026, 11:08
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Since it is sum of two integers numbers the maximum value of CDC <199, since first and third place integer is same , it will be 1 so C =1 , Now CDC will be in format of 1D1.
Now the unit digit is 1, When adding up AB+BA , the unit digit sum would be (A+B) and only 4 combination of Distinct unit digit can add up to 1 i.e (2,9) , (3,8), (4,7), (5,6). Since we are calculating A*B*C*D , one addition combination will served as one AB combination . For example - 29 and 92 would be one combination so like this 4 combination will exist and every combination will have different A*B*C*D.
Now the unit digit is 1, When adding up AB+BA , the unit digit sum would be (A+B) and only 4 combination of Distinct unit digit can add up to 1 i.e (2,9) , (3,8), (4,7), (5,6). Since we are calculating A*B*C*D , one addition combination will served as one AB combination . For example - 29 and 92 would be one combination so like this 4 combination will exist and every combination will have different A*B*C*D.
10 Jul 2026, 11:22
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even if we take 2 largest single digits eg. (& 9, addition = 18 i.e. tens unit can only be 1, hence 3 digit no. would start with 1, hence C=1 to get 1 in ones place as well, combos = A = 9, B = 3, A= 8, B = 4.....A=3, B=9. D = always 2. Hence unique combo of A*B*C*D = 6
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10 Jul 2026, 11:29
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IMO C
First of all, C can take only 1 value =1 (since A & B are non zero distinct integers, max value = 8+9 =17 so C will be the carry over 1)
Now working right to left, unit digit: A+B =C =1 , it should have a carry over so that D= A+B+1 else D will be equal to C, but Q says all are distinct
So clearly A & B should have values such that A+B= 11
So Possible values of pair (A,B) = (2,9 or 9,2), (3,8 or 8,3), (4,7 or 7,4), (5,6 or 6,5)
Using all the values every time we get D =2
Hence, diff values for A*B*C*D we get =4
Hope this helps
First of all, C can take only 1 value =1 (since A & B are non zero distinct integers, max value = 8+9 =17 so C will be the carry over 1)
Now working right to left, unit digit: A+B =C =1 , it should have a carry over so that D= A+B+1 else D will be equal to C, but Q says all are distinct
So clearly A & B should have values such that A+B= 11
So Possible values of pair (A,B) = (2,9 or 9,2), (3,8 or 8,3), (4,7 or 7,4), (5,6 or 6,5)
Using all the values every time we get D =2
Hence, diff values for A*B*C*D we get =4
Hope this helps
Bunuel
