Question

The total number of two digit numbers ‘$n$’, such that $3^n+7^n$ is a multiple of $10$, is

Answer 1

Step 1. Find the $n$

The given expression is $3^n+7^n$

$=3^n+{\left(10+\left(-3\right)\right)}^n$

Now applying binomial theorem

$\left[{(\mathrm{x}+\mathrm{y})}^{\mathrm{n}}=x^{\mathrm{n}}+\left(\begin{array}{c}\mathrm{n}\\ 1\end{array}\right)x^{\mathrm{n}-1}y+\left(\begin{array}{c}\mathrm{n}\\ 2\end{array}\right)x^{\mathrm{n}-2}{y}^2+...+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{n}-1\end{array}\right)x{y}^{\mathrm{n}-1}+y^{\mathrm{n}}\right]$

$$\begin{gathered} 3^n+{\left(10+\left(-3\right)\right)}^n \\ =3^n+{10}^n+\left(\begin{array}{c}n\\ 1\end{array}\right){10}^{n-1}\left(-3\right)+\left(\begin{array}{c}n\\ 2\end{array}\right){10}^{n-2}{\left(-3\right)}^2+...+\left(\begin{array}{c}n\\ n-1\end{array}\right)10·{\left(-3\right)}^{n-1}+{\left(-3\right)}^n \\ =\left\{\begin{array}{l}{10}^n+\left(\begin{array}{c}n\\ 1\end{array}\right){10}^{n-1}\left(-3\right)+\left(\begin{array}{c}n\\ 2\end{array}\right){10}^{n-2}{\left(-3\right)}^2+...+\left(\begin{array}{c}n\\ n-1\end{array}\right)10·{\left(-3\right)}^{n-1}\mathrm{when}n\mathrm{is}\mathrm{odd}\\ 2·3^n+{10}^n+\left(\begin{array}{c}n\\ 1\end{array}\right){10}^{n-1}\left(-3\right)+\left(\begin{array}{c}n\\ 2\end{array}\right){10}^{n-2}{\left(-3\right)}^2+...+\left(\begin{array}{c}n\\ n-1\end{array}\right)10·{\left(-3\right)}^{n-1}\mathrm{when}n\mathrm{is}\mathrm{even}\end{array}\right. \end{gathered}$$

Therefore clearly when $n$is odd then it is multiple of $10$

Step 2. Find the total number of two digit number

Clearly $n$ is odd

The first two digit such odd is $11$

Therefore $a_1=11$

The second two digit such odd is $13$

Therefore $a_n=13$

So common difference $d=2$

The last such digit is $a_N=99$

Therefore we know that the last term of an A.P series is represented by $a_N=a_1+\left(N-1\right)d$

$$\begin{gathered} \therefore a_N=a_1+\left(N-1\right)d \\ \Rightarrow 99=11+2\left(N-1\right) \\ \Rightarrow 88=2\left(N-1\right) \\ \Rightarrow 44=N-1 \\ \Rightarrow N=45 \end{gathered}$$

Hence, there are $45$ two digit number