First slide

Theorems of probability

Question

If the integer s 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 equals

Difficult

A

$\frac{1}{4}$
B

$\frac{1}{7}$
C

$\frac{1}{8}$
D

$\frac{1}{49}$

Solution

We know that $7^{1} = 7, 7^{2} = 49, 7^{3} = 343, 7^{4} = 2401, 7^{5} = 16807$ . Therefore $7^{k}$ (where $k \in Z$ ) results in a number whose unit's digit is 7 or 9 or 3 or 1.
Now $7^{m} + 7^{n}$ will be divisible by 5 if unit's place digit in the 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
$\begin{array}{l} m = 1, 5, 9, .... 97 \\ n = 3, 7, 11, ...., 99 \end{array}\}$ (25 options each) [7+3=10]
or
$\begin{array}{l} m = 2, 6, 10, ...., 98 \\ n = 4, 8, 12, ...., 100 \end{array}\}$ (25 options each) [9+3=10]
Therefore, the total number of selections of m, n such that $7^{m} + 7^{n}$ is divisible by 5 is $\frac{(25 \times 25 + 25 \times 25)}{100 \times 100} = \frac{1}{4}$

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