Question

If \(n\) is any natural number then \(6^n - 5^n\) always ends with :

• A. 1
• B. 3
• C. 5
• D. 7

Hint:

1. Last digit of \(6^n\) (for \(n \ge 1\)): \(6^1=6, 6^2=36, 6^3=216, \ldots\) The last digit is always 6. 2. Last digit of \(5^n\) (for \(n \ge 1\)): \(5^1=5, 5^2=25, 5^3=125, \ldots\) The last digit is always 5. 3. We need the last digit of (\text{a number ending in 6}) - (\text{a number ending in 5}). This will be a number ending in \(6-5 = 1\). Example for \(n=1\): \(6-5=1\). Example for \(n=2\): \(36-25=11\) (ends in 1).

Answer 1

The Correct Option is A

Concept: We need to find the last digit (unit digit) of the expression \(6^n - 5^n\) for any natural number \(n\). This involves looking at the patterns of the last digits of powers of 6 and 5. Step 1: Pattern of the last digit of powers of 6
\(6^1 = 6\)
\(6^2 = 36\)
\(6^3 = 216\)
\(6^4 = 1296\) The last digit of any positive integer power of 6 is always 6. So, the last digit of \(6^n\) is 6 for any natural number \(n\). Step 2: Pattern of the last digit of powers of 5
\(5^1 = 5\)
\(5^2 = 25\)
\(5^3 = 125\)
\(5^4 = 625\) The last digit of any positive integer power of 5 is always 5. So, the last digit of \(5^n\) is 5 for any natural number \(n\). Step 3: Find the last digit of \(6^n - 5^n\) We are interested in the last digit of the result of the subtraction. Let L(N) denote the last digit of a number N. We need L(\(6^n - 5^n\)). This is equivalent to L( L(\(6^n\)) - L(\(5^n\)) ), considering potential borrowing if L(\(6^n\)) < L(\(5^n\)). We have: Last digit of \(6^n\) is 6. Last digit of \(5^n\) is 5. So, we are looking at the last digit of a number ending in 6 minus a number ending in 5. Example: \( \ldots 6 - \ldots 5 \) The subtraction will result in a number ending in \(6 - 5 = 1\). For instance:
If \(n=1\): \(6^1 - 5^1 = 6 - 5 = 1\). (Last digit is 1)
If \(n=2\): \(6^2 - 5^2 = 36 - 25 = 11\). (Last digit is 1)
If \(n=3\): \(6^3 - 5^3 = 216 - 125 = 91\). (Last digit is 1) Since \(6^n\) will always be greater than \(5^n\) for natural number \(n\), no borrowing issues affect the simple subtraction of the last digits. The last digit of \(6^n - 5^n\) is always 1.