Question

The number 'a' is randomly selected from the set $\{0,1,2,3,..98,99\}$. The number 'b' is selected from the same set. Probability that the number $3^a+7^b$ has a digit equal to $8$ at the units place, is?

• A. $\frac{1}{16}$
• B. $\frac{2}{16}$
• C. $\frac{4}{16}$
• D. $\frac{3}{16}$

Answer 1

Answer: (D)

$3^a$ can have last digit of 3, 9, 7, 1

$7^b$ can have last digit of 7, 9, 3, 1

$3^a+7^b$ has 8 at units place if $3^a$ ends with 7, $7^b$ with 1

$3^a$ with 9, $7^b$ with 9

$3^a$ with 1, $7^b$ with 7

i) $3^a$ ends with 7 for 4n+3 from

$7^b$ ends with 1 for 4n from

So, probability $=\frac{25C_1.25C_1}{100\times 100}$

$=\frac{25\times 25}{100\times 100}$

$=\frac{1}{16}$

ii, iii) similarly for other 2 cases also, the probability $=\frac{1}{16}$

So, Total probability $=\frac{1}{16}+2\times \frac{1}{16}=\frac{3}{16}.$