Question

If the integers ${}^{\prime }{m}^{\prime }$ and ${}^{\prime }{n}^{\prime }$ are chosen at random from $1$ to $100$ then the probability that $7^m+7^n$ is divisible by $5$ is ?

• A. $1/5$
• B. $1/7$
• C. $1/4$
• D. $1/49$

Answer 1

Answer: (C)

$1/4$

$7^m$ can be 7,49,343,2401,16807.....

Last digits are

$7^{4m}$=1, $7^{4m+1}$=7,$7^{4m+2}$=9,$7^{4m+3}$=3

Therefore ,$7^m$ has last digits 7,9,3,1

For $7^m$+$7^n$ to be divisible by 5 last digit must be combination of 7,3 and 9,1.

Total possible way to choose m and n =100*100

There are 25 types of 4m between 1 to 10025 types of 4m+1 between 1 to 100,

25 types of 4m+2 between 1 to 100,25 types of 4m+3 between 1 to 100

So ways to choose m and n such that last digits are 7,3 or 3,7=2*25*25

So ways to choose m and n such that last digits are 1,9 or 9,1=2*25*25

Total ways =4*25*25

Probability=4*25*25/(100*100)=1/4