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?