Question

Let N be a natural number. If its first digit (from the left) deleted, it gets reduced to `(N)/(29)`. The sum of all the digits of N is
A. 14
B. 17
C. 23
D. 29

Answer 1

Correct Answer - A
Let `N=a_(n)a_(n-1)a_(n-2) . . .a_(3)a_(2)a_(1)a_(0)`
`=a_(0)+10a_(1)+10^(2)a_(2)+ . . .+10^(n-1)a_(n-1)+10^(n)a_(n)` . . . (i)
Then, `(N)/(29)=a_(n-1)a_(n-2)a_(n-3) . . .a_(3)a_(2)a_(1)a_(0)`
`=a_(0)+10a_(1)+10^(2)a_(2)+ . . .+10^(n-2)a_(n-2)+10^(n-1)a_(n-1)`
or `N=29(a_(0)+10a_(1)+10^(2)a_(2)+ . . .+10^(n-1)a_(n-2)+10^(n-1)a_(n-1))` . . . (ii)
From eqs. (i) and (ii), we get
`10^(n)*a_(n)=28(a_(0)+10a_(1)+10^(2)a_(2)+ . . .+10^(n-1)a_(n-1))`
`implies28` divided `10^(n)*a_(n)impliesa_(n)=7,n ge2 implies5^(2)=a_(0)+10a_(1)`
The required N is 725 orr 7250 or 72500, etc.
`therefore`The sum of the digits is 14.