Mathematics - Probability Question with Solution | TestHub

If the integers $m$ and $n$ are chosen at random from $\{1,2,3,\dots \dots ,100\}$ then the probability that a number of the form $7^m+7^n$ is divisible by 5 is equal to $\frac{1}{k}$. The numerical value of $k$ is

$$\begin{array}{rlrlrl}& 7^5=2k+2,7^6=5k+4,7^7=5k+3,7^8=5k+1& \therefore 7^1,7^5,7^9,\dots \dots 7^{97}\text{ are of the type }5k+2& 7^2,7^6,7^{10},\dots \dots 7^{98}\text{ are of the type }5k+4& 7^3,7^7,7^{11},\dots \dots 7^{99}\text{ are of the type }5k+3& \&7^4,7^8,7^{12},\dots \dots 7^{100}\text{ are of the type }5k+1\end{array}$$

There will be only four favourable combinations in which $7^m+7^n$ will be divisible by 5 .

$$\begin{array}{rlr}& 7^m:5k+2,5k+4,5k-3,5k+1& 7^n:5k+3,5k+1,5k+2,5k+4\end{array}$$

∴ Number of favourable cases is

$$n(E)=25\times 25+25\times 25+25\times 25+25\times 25=4\times 625$$

$$\text{ ∴ }\text{ Required probability }=\frac{n(E)}{n(S)}=\frac{4\times 625}{100\times 100}=\frac{1}{4}$$