KA-0149Last-digit behavior of products and powers
4 points, difficulty 2 of 3Level 5–6about 90sWhat is the units digit of 3 multiplied by itself 100 times, that is 3 to the power 100?
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
The last digit of a product depends only on the last digits being multiplied. Work out the last digits of the first few powers until the list repeats, then locate 100 inside that cycle.
3The full solution
The last digits run 3, 9, 7, 1 and then repeat. 100 divides exactly by 4, so it lands on the last entry of a cycle, which is 1.
Solution
The reliable way
Build the cycle of last digits: 3, then 9, then 27 ends in 7, then 81 ends in 1, then it starts again at 3. The cycle has length 4. Now 100 divided by 4 is 25 with remainder 0, and a remainder of 0 means the last entry of a cycle, not the first - so the answer is 1. The transferable idea: find the cycle, then use the remainder as a position inside it, remembering that remainder zero lands at the end.
The elegant way
3 to the power 4 is 81, which ends in 1, and 100 is a multiple of 4 - so the answer ends in 1.
Why this is on the test: The remainder-of-zero case is the one students get wrong, and choices C and D are the two ways of being one step out.