If the integers m and n are chosen at random between 1 and 100, then the proba

Question

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 equals

Options:

A .

\frac{1}{4}

B .

\frac{1}{7}

C .

\frac{1}{8}

D .

\frac{1}{49}

Answer: Option A
:
A
Since m and n are selected between 1 and 100, hence sample space =

100 \times 100.

Also

7^{1} = 7, 7^{2} = 49, 7^{3} = 343, 7^{4} = 2401, 7^{5} = 16807

etc. Hence 1, 3, 7 and 9 will be the last digits in the powers of 7. Hence for favorable cases

n m \to ↓ 1, 11, 21, 3.... 1, 100 2, 12, 22, 3... 2, 100 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100, 1100, 2100, 3100, 100

For m = 1; n = 3, 7, 11….. 97

∴

Favorable cases = 25
Similarly for every m, favorable n are 25.

∴

Total favourable cases =

100 \times 25

Hence required probability =

\frac{100 \times 25}{100 \times 100} = \frac{1}{4}

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