Math CentralQuandaries & Queries


Question from Shankar, a teacher:
Can the cube of an integer end with 1985 ones?

Can the cube of an integer end with one 1?

Can it end with 2 ones? three 1's? (Experiment)

We can show by induction that there always exists a number ending in k 1's.

Suppose N < 10k-1 and N3 ends in at least (k-1) 1's. Then there exists A, 0<=A<10, such that A*10k-1 + N3 ends with k 1's .

But there exists (prove this!) B with 0<=B<10 such that 3B ends in A. Show that [B*10k-1 + N]3 ends in k 1's. Hint: binomial expansion.

Bonus question: why does this NOT work for squares? What is the next power it does work for?

Good Hunting!

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