Firstly let us solve for unit’s place,

B + 1 = 8

B = 7

Now let us solve for ten’s place,

A + B = 1

A + 7 = 1

A = -6 which is not possible.

Hence, A + B > 9

We know that now there should be one carry in hundred’s place and so we need to subtract 10 from ten’s place,

i.e., A + B – 10 = 1

A + 7 = 11

A = 11-7 = 4

Now to check whether our values of A and B are correct, we should solve for hundred’s place.

2 + A + 1 = B

2 + 4 + 1 = 7

7 = 7

i.e., RHS = LHS

A = 4 and B = 7

Answered by Aaryan | 1 year agoWhich of the following numbers are divisible by 11:

**(i) **10835

**(ii)** 380237

**(iii)** 504670

**(iv) **28248

In each of the following replace * by a digit so that the number formed is divisible by 11:

**(i) **64 × 2456

**(ii) **86 × 6194

In each of the following replace * by a digit so that the number formed is divisible by 6:

**(i)** 97 × 542

**(ii)** 709 × 94

In each of the following replace × by a digit so that the number formed is divisible by 9

**(i) **49 × 2207

**(ii) **5938 × 623

If 42z3 is a multiple of 9, where z is a digit, what is the value of z?