Question
If the integers $$m$$ and $$n$$ are chosen at random from 1 to 100, then the probability that a number of the form $${7^m} + {7^n}$$ is divisible by 5 equals
A.
$$\frac{1}{4}$$
B.
$$\frac{1}{7}$$
C.
$$\frac{1}{8}$$
D.
$$\frac{1}{49}$$
Answer :
$$\frac{1}{4}$$
Solution :
We know that,
$${7^1} = 7,{7^2} = 49,{7^3} = 343,{7^4} = 2401,{7^5} = 16807$$
∴ $${7^k}$$ (where $$k \in Z$$ ), results in a number whose unit’s digit 7 or 9 or 3 or 1.
Now, $${7^m} + {7^n}$$ will be divisible by 5 if unit’s place digit of
resulting number is 5 or 0 clearly it can never be 5.
But it can be 0 if we consider values of $$m$$ and $$n$$ such that the sum of unit’s place digits become 0. And this can be done by choosing
\[\begin{array}{l}
\left. \begin{array}{l}
m = 1,5,9,.....97\\
{\rm{and \,correspondingly}}\\
n = 3,7,11,.....99
\end{array} \right\}\,\,\left( {25\,{\rm{options\, each}}} \right)\left[ {7 + 3 = 10} \right]\\
\left. \begin{array}{l}
m = 2,6,10,.....98\\
{\rm{and}}\\
n = 4,8,12,.....100
\end{array} \right\}\,\,\left( {25\,{\rm{options \,each}}} \right)\left[ {9 + 1 = 10} \right]
\end{array}\]
Case I : Thus $$m$$ can be chosen in 25 ways and $$n$$ can be chosen in 25 ways
Case II : $$m$$ can be chosen in 25 ways and $$n$$ can be chosen in 25 ways
∴ Total no. of selections of $$m, n$$ so that $${7^m} + {7^n}$$ is divisible by $$5 = \left( {25 \times 25 + 25 \times 25} \right) \times 2$$
Note we can interchange values of $$m$$ and $$n.$$
Also no. of total possible selections of $$m$$ and $$n$$ out of $$100 = 100 \times 100$$
∴ Req. prob. $$ = \frac{{2\left( {25 \times 25 + 25 \times 25} \right)}}{{100 \times 100}} = \frac{1}{4}.$$