Aptitude - Numbers - Discussion

How can you got (9 - b) I cannot understand. Please, Explain

-> (7 + 3) - (b + 1)

simplify this u got (9 - b)

I could not get why 13b7 divisible 11 was represented as (7+3) - (b+1). ?

Soma-->
the number is divisible by 11 if the difference between the sum of its digit at odd places and the sum of its digit at even places is either 0 or number divisible by 11.
4 a 3 | ==> here a can either be 0 or 1
9 8 4 }
13 b 7 |==> so b can be either 8 or 9

but (7+3) - (b+1) should be divisible by 11 or 0
so b can't be 8. so b is 9 and a is 1.
hence (a+b) is 10.

I could not understand this problem.

i.e. How we got (7+3) - (b+1) from 13b7 ?

13b7 to know the divisibility of 11, we have to subtract sum of odd placed numbers with sum of even placed numbers.

in 13b7 odd places are 7and3

even places are b and 1.

now got it.

How do we get (9 - b) = 0 Please explain.....

->211* 11=2321
now (2+2)-(3+1)=0.
simplify u got (9 - b)=0

Guys i'm confused, how com we take a+8, is there simple method?

If we add 4a3 to 984 then we can clearly see that the result is such a way that the digits in 100'ths place have no carry.

That means a+8 is always less than 9 in order to get 13b7 by comparing the places in addition the 10'ths digits underwent.

a+8=b --- (1).

Then 13b7 is divisible by 11 so the difference of sets, sum of even places and odd places in 13b7 is.

1+b-10=0 or multiple of 11.

Try 0.

Then b=9 which is less or equal to 9 hence by eq (1).

We can get answer.

If you try other than 11 then this adds up a carry in first addition of two number.