Solution:
Now,
For A + 5, we get unit digit 2
For A = 7, we get 7 + 5 = 12
Hence, the value of B carried on 1
Means, +3 + 2 = 6
Hence, the value of A = 7 and B = 6
Solution:
For A + 8 , we get unit digit 3.
For 8 + 5 = 13 in which 3 is at ones place.
Therefore, A = 5 and carry over 1 then
B = 4 and C = 1
Hence, A = 5, B = 4 and C = 1
Solution:
A x A = A
Now, for A = 1 or 6
For 1 x 1 = 1 which is not equal to 9
For, A = 6
We get, 6 x 6 = 36
So, A x 1 = 6 x 1 + 3 (carried on) = 9
Therefore, A = 6
Solution:
3 + A = 6
A = 6 – 3 = 3
So, 2 + 1 is carried on
A = 2
Now,
B + 7 = number must have unit digit 2 i.e 12
So, B = 12 – 7 = 5
Hence A = 2 and B =5
Solution:
In given we can put 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
Solution:
When we put B = 0,
we get 0 x 5 = 0 and A = 5,
Then, 5×5 =25
A = 5, C = 2
Hence A = 5, B = 0 and C =2
Solution:
According to given question the product of B and 6 must be same as ones place digit as B.
B x 6 = B
Now,
6×1 = 6,
6×2 = 12,
6×3 = 18,
6×4 =24
So,
On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44.
Therefore, for 6×7 = 42+2 =44
Hence A = 7 and B = 4
Solution:
1 + B = 0
On putting B = 9, we get 9+1 = 10
Now, Putting 0 at ones place and carry over 1, we get for A = 7
7+1+1 =9
Hence, A = 7 and B = 9
Solution:
B + 1 = 8
B = 8 -1
B = 7
Foe A + B = a number must have unit digit 1
A + B =11
A + 7 = 11
A = 11 – 7 = 4 (1 carried to)
Now, 1 carried on + 2 + A = B
3 + A = 7
A = 7 – 3 = 4
Hence, A = 4 and B = 7
Solution:
A + B = 9
Means, 9 = A + B
9 = 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5 or 0 + 9 or 9 + 0
Now, for 2 + A , 0 is required at unit place
2 + A = 10,
A = 10 – 2 = 8
B = 9 – 8 = 1
1 + 6 + 1 (carried on) = A = 8
Hence A = 8 and B =1
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