Algebra Arithmetic Math Olympiad Math Olympiad Videos

Divisibility Problem | HANOI 2018

Try this beautiful problem from HANOI, 2018 based on Divisibility. You may use sequential hints to solve the problem if required.

Try this beautiful problem from American Invitational Mathematics Examination, HANOI, 2018 based on Divisibility.

Divisibility – HANOI 2018

Find all positive integers k such that there exists a positive integer n, for which \(2^{n}+11\) is divisible by \(2^{k}-1\).

  • is 2, 4
  • is 2, 0
  • is 1,2,4
  • cannot be determined from the given information

Key Concepts

Number Theory



Check the Answer

But try the problem first…

Answer: is 1,2,4.

Suggested Reading

HANOI, 2018

Elementary Number Theory by David Burton

Try with Hints

First hint

Suppose n is a positive integer such that \(2^{n}+11 \) is divisible by \(2^{k}-1\). Let n=qk+r where q,r are nonnegetive integers and \( 0 \leq r \lt k\). We have \(2^{n}+11=2^{kq+r}+11=2^{r}(2^{kq}-1)+(2^{r}+11)\) in comparision with \( 2^{k}-1\) implies \(2^{r}+11 \) in comparision with \(2^{k}-1\).

Second Hint

Thus we have \(2^{r}+11 \geq 2^{k}-1\) and \(r \lt k\) Then \(k \leq 4\).

Final Step

By calculating for k=1,2,3,4 we obtain k=1,2,4..

Subscribe to Cheenta at Youtube

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.