Understanding the Problem: Last Digit of \(6^n - 5^n\)

The question asks us to determine the last digit of the expression \(6^n - 5^n\), where \(n\) is any natural number. Natural numbers are \(1, 2, 3, \dots\).

To find the last digit of an expression involving powers, we only need to focus on the last digits of the base numbers involved in the calculation.

Exploring Last Digits of Powers of 6 and 5

Let's look at the pattern of the last digits of powers of 6 and 5 for natural numbers \(n \ge 1\).

Last Digit Pattern of \(6^n\)

• For \(n=1\), \(6^1 = 6\). The last digit is 6.
• For \(n=2\), \(6^2 = 36\). The last digit is 6.
• For \(n=3\), \(6^3 = 216\). The last digit is 6.
• For \(n=4\), \(6^4 = 1296\). The last digit is 6.

It is clear that for any natural number \(n \ge 1\), the last digit of \(6^n\) is always 6.

Last Digit Pattern of \(5^n\)

• For \(n=1\), \(5^1 = 5\). The last digit is 5.
• For \(n=2\), \(5^2 = 25\). The last digit is 5.
• For \(n=3\), \(5^3 = 125\). The last digit is 5.
• For \(n=4\), \(5^4 = 625\). The last digit is 5.

It is clear that for any natural number \(n \ge 1\), the last digit of \(5^n\) is always 5.

Calculating the Last Digit of the Difference

We need to find the last digit of \(6^n - 5^n\). Since the last digit of \(6^n\) is always 6 and the last digit of \(5^n\) is always 5 (for \(n \ge 1\)), the last digit of their difference will be the last digit of \(6 - 5\).

The difference of the last digits is \(6 - 5 = 1\).

This means that regardless of the specific value of \(n\) (as long as it's a natural number, i.e., \(n \ge 1\)), the subtraction \(6^n - 5^n\) will always result in a number whose last digit is 1.

Illustrative Examples for \(6^n - 5^n\)

Let's verify with a few values of \(n\):

• For \(n=1\): \(6^1 - 5^1 = 6 - 5 = 1\). Last digit is 1.
• For \(n=2\): \(6^2 - 5^2 = 36 - 25 = 11\). Last digit is 1.
• For \(n=3\): \(6^3 - 5^3 = 216 - 125 = 91\). Last digit is 1.

These examples confirm that the last digit is consistently 1.

Conclusion on the End Digit

For any natural number \(n\), \(6^n\) always ends in 6 and \(5^n\) always ends in 5. Subtracting a number ending in 5 from a number ending in 6 results in a number ending in 1. Therefore, \(6^n - 5^n\) always ends with 1.

Revision Table: Key Concepts for Last Digits

Additional Information: Modular Arithmetic and Unit Digits

Finding the last digit is equivalent to finding the remainder when a number is divided by 10. This concept is formally studied in modular arithmetic.

• The last digit of a number \(X\) is \(X \pmod{10}\).
• For powers, we look at the pattern of \(b^n \pmod{10}\) where \(b\) is the base.
• For \(6^n \pmod{10}\): \(6^1 \equiv 6 \pmod{10}\), \(6^2 = 36 \equiv 6 \pmod{10}\), \(6^3 = 216 \equiv 6 \pmod{10}\). For \(n \ge 1\), \(6^n \equiv 6 \pmod{10}\).
• For \(5^n \pmod{10}\): \(5^1 \equiv 5 \pmod{10}\), \(5^2 = 25 \equiv 5 \pmod{10}\), \(5^3 = 125 \equiv 5 \pmod{10}\). For \(n \ge 1\), \(5^n \equiv 5 \pmod{10}\).
• We want to find \((6^n - 5^n) \pmod{10}\).
• In modular arithmetic, \((a - b) \pmod{m} \equiv (a \pmod{m} - b \pmod{m}) \pmod{m}\).
• So, \((6^n - 5^n) \pmod{10} \equiv (6^n \pmod{10} - 5^n \pmod{10}) \pmod{10}\).
• \((6^n - 5^n) \pmod{10} \equiv (6 - 5) \pmod{10}\).
• \((6^n - 5^n) \pmod{10} \equiv 1 \pmod{10}\).

This confirms that the last digit of \(6^n - 5^n\) is 1 for any natural number \(n\).