Question

Number of points having position vector $ai+bi+ck,a,b,c\in \left\{1,2,3,4,5\right\}$ such that $2n+3b+5c$ is divisible by $4$ is?

• A. $140$
• B. $70$
• C. $100$
• D. $75$

Answer 1

The correct option is B

$70$

Explanation for the correct option.

Finding the Number of points having position vector $ai+bi+ck,a,b,c\in \left\{1,2,3,4,5\right\}$ such that $2n+3b+5c$ is divisible by $4$ is

Given: position vector $ai+bi+ck,a,b,c\in \left\{1,2,3,4,5\right\}$

Let, $4m=2^a+3^b+5^c$

$$\begin{gathered} 4m=2^a+{(4-1)}^b+{(4+1)}^c \\ =2^a+4k+{(-1)}^b+{\left(1\right)}^c\left[k\text{is constant}\right] \end{gathered}$$

If $a=1,b=even,c=anynumber$

If $a\ne 1,b=odd,c=anynumber$

Therefore, the required number of ways $=(1\times 2\times 5)+(4\times 3\times 5)=70$

Hence, option (B) is the correct answer.