If the integers $$m$$ and $$n$$ are chosen at random between $$1$$ and $$100$$  then the probability that a number of the form $${7^m} + {7^n}$$   is divisible by $$5$$ is :

Solution :
We know $${7^k},\,k\, \in \,N,$$   has $$1,\,3,\,9,\,7$$   at the units place for $$k = 4p,\,4p - 1,\,4p - 2,\,4p - 3$$       respectively, where $$p = 1,\,2,\,3,.....$$
Clearly, $${7^m} + {7^n}$$   will be divisible by $$5$$ if $${7^m}$$ has $$3$$ or $$7$$ in the units place and $${7^n}$$ has $$7$$ or $$3$$ in the units place or $${7^m}$$ has $$1$$ or $$9$$ in the units place and $${7^n}$$ has $$9$$ or $$1$$ in the units place.
$$\therefore $$  for any choice of $$m,\,n$$  the digit in the units place of $${7^m} + {7^n}$$   is $$2,\,4,\,6,\,0$$   or $$8.$$ It is divisible by $$5$$ only when this digit is $$0.$$
$$\therefore $$  the required probability $$ = \frac{1}{5}.$$