Discussion Forum

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

Answer:

Explanation:

let I=$7^{n}+7^{m}$  , then we observe that $7^{i}$ ,$7^{2}$ ,$7^{3}$  and $7^{4}$  ends in 7,9,3 and 1 respectively. Thus , $7^{i}$  ends in 7,9,3 or 1 according as  i  is of the form 4k+1,4k+2,4k-1 respectively 

 If  S  is the sample space, then $n(S) =(100)^{2}$

$7^{m}+7^{n}$ is divisible by 5, if 

(i)  m is of the form 4k+1 and n is of the form  4k-1 or

(ii) m is of the form 4k+2 an n is of the form 4k or 

(iii)    m is the form 4k-1 and n is of the form 4k+1 or 

(iv)   m is of the form 4k and n is of the form 4k+1 or 

 So, number of favourable  ordered pairs (m,n)  =$4 \times 25 \times 25$

$\therefore$   Required probability= $\frac{4 \times 25 \times 25}{(100)^{2}}= \frac{1}{4}$