KA-0158Last-digit behavior of products and powers
5 points, difficulty 3 of 3Level 7–8about 120sWhat is the last digit of 3 multiplied by itself 2026 times, that is, of 3 to the power 2026?
Hints
Take them one at a time. The first gives nothing away.
1A nudge
List the last digits of the first few powers.
2The strategy
3, 9, 27, 81, 243 end in 3, 9, 7, 1, 3 - the last digits repeat every four powers.
3The full solution
2026 divided by 4 leaves remainder 2, so 3 to the power 2026 ends like 3 squared, which is 9.
Solution
The reliable way
Last digits of powers always repeat. Listing them: 3^1 ends in 3, 3^2 in 9, 3^3 in 7, 3^4 in 1, and then 3^5 ends in 3 again. The cycle is 3, 9, 7, 1 and it has length 4. To find where 2026 lands, divide by 4: 2026 = 4 x 506 + 2, so the remainder is 2 and the answer matches the second entry of the cycle, which is 9. Watch the off-by-one: remainder 1 means the first entry, remainder 0 means the last. The transferable idea: every last-digit question is a cycle question, and the work is locating the exponent in the cycle.
The elegant way
3^4 ends in 1, so 3^2024 ends in 1 and two more threes take it to 9.
Why this is on the test: Every last-digit question is really a cycle question, and the marks are lost on where the remainder lands, not on finding the cycle.