Question

The number of points having position vectors $a\hat i + b\hat j + c\hat k$ where $a,b,c \in \left\{ {1,2,3,4,5} \right\}$ such that ${2^a} + {3^b} + {5^c}$ is divisible by $4$ is-
A.$70$
B.$140$
C.$210$
D.$280$

Answer 1

Hint: Use the formula ${\text{Dividend = divisor}} \times {\text{qoutient + remainder}}$ as it is given that ${2^a} + {3^b} + {5^c}$ is completely divisible by $4$ leaving no remainder. Solve the equation and use the formula of combination which is given as-
${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$ where n is the total number of things and r is the number of things to be selected to select the numbers for a, b and c from the set $\left\{ {1,2,3,4,5} \right\}$. Add the required ways to choose all the numbers to get the number of points.

Complete step by step answer:

Given the number of points have position vectors $a\hat i + b\hat j + c\hat k$ where $a,b,c \in \left\{ {1,2,3,4,5} \right\}$
Such that ${2^a} + {3^b} + {5^c}$ is divisible by $4$.
Now if ${2^a} + {3^b} + {5^c}$ is divisible by $4$ then its remainder will be zero. Let the quotient be m.
By the division formula,
$ \Rightarrow $ Divisor ×quotient + remainder= Dividend
On putting Dividend=${2^a} + {3^b} + {5^c}$ , divisor =$4$ and remainder= $0$ , we get,
$ \Rightarrow 4m + 0 = {2^a} + {3^b} + {5^c}$
$ \Rightarrow 4m = {2^a} + {3^b} + {5^c}$
Now we can write,${3^b} = {\left( {4 - 1} \right)^b}$ and ${5^c} = {\left( {4 + 1} \right)^c}$ as it will not change the value of the equation,
So on putting these in the equation, we get-
$ \Rightarrow 4m = {2^a} + {\left( {4 - 1} \right)^b} + {\left( {4 + 1} \right)^c}$
Here even if the value of b is odd or even, we would get all the multiples of $4$ on opening the brackets except ${\left( { - 1} \right)^b}$ and${\left( 1 \right)^c}$ .
So assuming them to be $4k$ we get,
$ \Rightarrow 4m = {2^a} + {\left( { - 1} \right)^b} + {1^c} + 4k$
Now here since $1$ has power c raised to it so c can be any number as it will always give the value $1$
Then we have to find a and b, Now here since$a,b,c \in \left\{ {1,2,3,4,5} \right\}$
Then a≠0, a can have value either equal to one or greater than one.
So condition (I) is-
When a=$1$ then b =even number and c= any number.
Now using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$ we can select the numbers for a, b and c from set$\left\{ {1,2,3,4,5} \right\}$
Since there are $2$ even number$\left( {2,4} \right)$ then the number of ways to select one of them for b =${}^2{C_1} = \dfrac{{2!}}{{1!1!}} = 2$
And number of ways to select any number for c=${}^5{C_1} = \dfrac{{5!}}{{4!1!}} = 5$
So number obtained from condition (I) = $1 \times 2 \times 5 = 10$ --- (i)
And Condition (II) is-
When a≠$1$ then b=odd number and c= any number.
So the number of ways of selecting one value of a from five values =${}^4{C_1}$ $ = 4$
The number of ways of selecting one odd number from given numbers for b=${}^3{C_1} = 3$
And number of ways to select any number for c=${}^5{C_1} = \dfrac{{5!}}{{4!1!}} = 5$
So number obtained from condition (II) =$4 \times 3 \times 5 = 60$ --- (ii)
So the required number= eq. (i) +eq. (ii)
Required number=$10 + 60 = 70$
Hence the correct answer is A.

Note: Here, since we know that ${2^a} + {3^b} + {5^c}$ is divisible by $4$ then values of a, b and c should be such that the remainder obtained is zero. Here we see that if a=$1$ then ${2^1} = 2$ then b has to be even number because then${\left( { - 1} \right)^{even}} = 1$ and we already know the value of c does not affect the number because ${1^n} = 1$ .Only in such a condition will the number be completely divisible by$4$. Also here the students may get confused as to why are we adding all the required number of ways to select the value of a, b, and c. We do so because we have to find the number of points (there can be $20$ such points or more than $20$such points) which satisfy the given condition. So the number of points will be equal to the number of ways the value of a, b and c can be selected from the given set.