1. Say, A = 7, and we get
7+5 = 12
In which one’s place is 2.
Therefore, A = 7
And putting 2 and carrying over 1, we get
B = 6
Hence, A = 7 and B = 6.
2. If A = 5, we get
8+5 = 13, in which one’s place is 3.
Therefore, A = 5 and carry over 1, then
B = 4 and C = 1
Hence, A = 5, B = 4 and C = 1.
3. On putting A = 1, 2, 3, 4, 5, 6, 7 and so on, we get
AxA = 6×6 = 36, in which one’s place is 6.
Therefore, A = 6
4. Here, we observe that B = 5, so that 7+5 =12
Putting 2 at one’s place and carrying over 1 and A = 2, we get
2+3+1 =6
Hence, A = 2 and B =5.
5. Here, on putting B = 0, we get 0x3 = 0.
And A = 5, then 5×3 =15
A = 5 and C = 1
Hence A = 5, B = 0 and C = 1.
6. On putting B = 0, we get 0x5 = 0 and A = 5, then 5×5 =25
A = 5, C = 2
Hence A = 5, B = 0 and C =2
7. Here, products of B and 6 must be the same as one’s place digit is B.
6×1 = 6, 6×2 = 12, 6×3 = 18, 6×4 = 24
On putting B = 4, we get the one’s digit 4, and the remaining two B’s value should be 44.
Therefore, for 6×7 = 42+2 =44
Hence, A = 7 and B = 4.
8. On putting B = 9, we get 9+1 = 10
Putting 0 at ones place and carrying over 1, we get A = 7
7+1+1 =9
Hence, A = 7 and B = 9.
9. On putting B = 7, we get 7+1 = 8
Now A = 4, then 4+7 = 11
Putting 1 at tens place and carrying over 1, we get
2+4+1 =7
Hence, A = 4 and B = 7.
10. Putting A = 8 and B = 1, we get
8+1 = 9
Now, again we add 2 + 8 =10
The tens place digit is ‘0’ and carries over 1. Now 1+6+1 = 8 = A
Hence, A = 8 and B =1.