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, is
Answer & explanation
Correct answer: option 1
We have,
Number of ways of selecting m and n = 100 × 100
We know that $7^k , k ∈ N $ has 1, 3, 9, 7, at the units place for
$k = 4λ, 4λ -1, 4λ -2, 4λ - 3.$
Therefore, $7^m + 7^n $ can have, 0, 2, 4, 6 and 8 at units place. But, $7^m + 7^n $ will be divisible by 5 if it has 0 at units place.
The digit at units place in $7^m + 7^n$ will be 0 if m and n have the following forms:
|
Digital at units place in m |
Form of m | Digit at units place in m | Form of n |
|
1 9 3 7 |
4λ 4λ-2 4λ-1 4λ -3 |
9 1 7 3 |
4λ-2 4λ 4λ-3 4λ -1 |
Clearly, for each value of m there are 25 values of n for which the digit at units place in $7^m + 7^n $ is 0.
Number of ways of selecting m and n so that $7^m + 7^n$ is divisible by 5 is
$ 25×25 + 25× 25 + 25 × 25 + 25 × 25 = 2500$
Hence, required probability $=\frac{2500}{100×100}=\frac{1}{4}$