Question

What is the unit's digit of the expression \(77^{920} + 64^{165} + 53^{246}\)?

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Use the repeating unit-digit cycle of each base to find the last digit of each power separately.

Answer 1

The Correct Option is A

Step 1: Find the unit-digit cycle of \(77^{920}\).
Only the unit digit of the base matters, and 77 ends in 7.
Powers of 7 end in 7, 9, 3, 1 and then repeat, a cycle of length 4.
Since \(920 = 4 \times 230\) is exactly divisible by 4, \(77^{920}\) ends in the fourth value of the cycle, which is 1.

Step 2: Find the unit-digit cycle of \(64^{165}\).
64 ends in 4, and powers of 4 end in 4, 6 and then repeat, a cycle of length 2.
Since 165 is odd, \(64^{165}\) ends in the first value of the cycle, which is 4.

Step 3: Find the unit-digit cycle of \(53^{246}\).
53 ends in 3, and powers of 3 end in 3, 9, 7, 1 and then repeat, a cycle of length 4.
Dividing 246 by 4 leaves a remainder of 2, so \(53^{246}\) ends in the second value of the cycle, which is 9.

Step 4: Add the three unit digits.
The three last digits are 1, 4 and 9, and \(1+4+9=14\), whose own unit digit is 4.

Step 5: Note on the answer key.
This cyclicity check gives a unit digit of 4, which was option (e) in the original five-choice paper.
The official answer key marks option (a), the value 0, as correct, so this row is flagged for review and the keyed option is kept unchanged as instructed.

Final Answer:
As per the official answer key the marked option is (a). \[ \boxed{0 \text{ (keyed option a)}} \]