First slide

Permutations

Question

The number of ordered pairs $(m, n), m, n$ $\in {1, 2, \dots, 100}$ such that $7^{m} + 7^{n}$ is divisible by 5 is

Moderate

A

1250
B

2000
C

2500
D

5000

Solution

Note that $7^{r} (r \in N)$ ends in $7, 9, 3$ or $1$ (corresponding to $r = 1, 2, 3$ and $4$ respectively.)
Thus $7^{m} + 7^{n}$ cannot end in 5 for any values of $m, n \in N .$
In other words, for $7^{m} + 7^{n}$ to be divisible by 5, it should end in 0.
For $7^{m} + 7^{n}$ to end in $0$ , the forms of m and $n$ should be as follows:
m n
1 4r 4s + 2
2 4r + 1 4s + 3
3 4r + 2 4s
4 4r + 3 4s + 1
Thus, for a given value of m there are just $25$ values of n for
which $7^{m} + 7^{n}$ ends in 0. [For instance, if $m = 4 r$ then $n =$ 2, 6, 10, …, 98]
$∴$ there are $100 \times 25 = 2500$ ordered pairs $(m, n)$ for
which $7^{m} + 7^{n}$ is divisible by 5.

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