**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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