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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

Inequality

Algebra

Check the Answer


But try the problem first…

Answer: is 1,2,4.

Source
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..

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