Cheenta
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

Divisibility Problem | HANOI 2018

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


Answer: is 1,2,4.

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


Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com