Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
Q1) Given a b c and d are distinct positive digits
AB + BA = CD4
which of the following can be value of A?
a) 1
b) 5
c) 7
d) 8
e) 9
Let's consider only unit digit. Then we have only two cases, namely B+A=4 or B+A=14 (since A and B are digit numbers they are less than or equal to 9. That means B+A should be less than or equal to 18. So B+A cannot be 24 neither 34,... ). If B+A=4, since A+B is also 4, AB+BA=44(contradiction to the assumption). So B+A should be 14. That is AB+BA= (A+B)*10 + (B+A) = 154. That means C=1 and D=5.
Now let's find out what A and B are. Since A+B= 14, we have only tree cases to make 14, namely A+B= 7+7, 8+6 or 9+5. But A, B, C, D should be distinct. So A+B=8+6 (since the cases A=7, B=7 and B=5, D=5 have same digits). A can be, therefore, 8 or 6. So the answer is D.
Q2) Given a b and C are distinct positive digits
AB + BA = CDC
what can be the values of C and D?
Similarly with Q1, B+A=C or B+A=1C. If B+A=C, similarly with Q1, A+B=C ----> AB+BA=CC(contradiction to the assumption). So B+A=1C. That means AB+BA= (1C)*10 + 1C. So 1 is a carrying to the tenth digit. That means C=1 and D=C+1=2. So the answer is C=1, D=2.
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