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

• A. $\frac{1}{4}$
• B. $\frac{1}{7}$
• C. $\frac{1}{8}$
• D. $\frac{1}{49}$

Hint:

Always check cycle of powers in modulo problems.

Answer 1

The Correct Option is A

To determine the probability that the number of the form \(7^m + 7^n\) is divisible by 5, we first need to understand the behavior of the powers of 7 modulo 5. This is because the result depends on the remainder when dividing by 5.

The powers of 7 modulo 5 exhibit a periodic pattern. Let's explore the first few powers:

• \(7^1 \equiv 2 \pmod{5}\)
• \(7^2 \equiv 4 \pmod{5}\)
• \(7^3 \equiv 3 \pmod{5}\)
• \(7^4 \equiv 1 \pmod{5}\)

The pattern repeats every four terms because:

• \(7^5 \equiv 7^1 \equiv 2 \pmod{5}\)

From the sequence above, we observe that powers of 7 modulo 5 cycle through the values \(2, 4, 3, 1\).

Now, for \(7^m + 7^n \equiv 0 \pmod{5}\), it is necessary that:

• \(7^m \equiv -7^n \equiv (5 - 7^n) \pmod{5}\)

This implies \(7^m + 7^n \equiv 0 \pmod{5}\), i.e., their respective cycle positions add up to a multiple of 5.

Considering the cycle positions:

• \(2 + 3 \equiv 5 \equiv 0 \pmod{5}\)
• \(3 + 2 \equiv 5 \equiv 0 \pmod{5}\)
• \(1 + 4 \equiv 5 \equiv 0 \pmod{5}\)
• \(4 + 1 \equiv 5 \equiv 0 \pmod{5}\)

The list shows four pairs that satisfy the divisibility condition: \((2,3), (3,2), (1,4), (4,1)\).

Since the cycle has four possible values and is repeated, the probability of selecting one of these pairs is \(\frac{4}{16} = \frac{1}{4}\).

This is because we consider all possible pairings of cycle positions, and only these four combinations satisfy the divisibility condition.

Therefore, the probability that \(7^m + 7^n\) is divisible by 5 is \(\frac{1}{4}\).