Question

Consider the number N=123times43y. If N is divisible by 9 then find the sum of all possible values of x+y solve this question here help ​

Answer 1

Correct Question:-

Consider the number N=123x43y. If N is divisible by 9 then find the sum of all possible values of x+y.

Solution:-

If the number N=123x43y is divisible by 9 then the sum of digits of N must be divisible by 9, i.e., for some positive integer n,

$\small\longrightarrow 1+2+3+x+4+3+y=9n$

$\small\longrightarrow x+y+13=9n$

$\small\longrightarrow x+y=9n-13$

Here, as x and y are digits, 0 ≤ x ≤ 9 and 0 ≤ y ≤ 9, so possibly 0 ≤ x+y ≤ 18, i.e.,

$\small\longrightarrow 0\leq9n-13\leq18$

$\small\longrightarrow 13\leq9n\leq31$

$\small\longrightarrow\dfrac{13}{9}\leq n\leq\dfrac{31}{9}$

As n is a positive integer,

$\small\longrightarrow n\in\{2,\ 3\}$

$\small\Longrightarrow x+y=9n-13\in\{9\times2-13,\ 9\times3-13\}$

$\small\longrightarrow x+y\in\{5,\ 14\}$

Hence the sum of all possible values of x+y is 5+14=19.