Problem. Determine the number of solutions of each following equation

  • $\sin v= \dfrac{v}{1964}$
  • $\sin v= \log_{100} v$

Remark. This problem is similar to another one in Moscow Olympiad 1941.

  • Remark: $v= 0$ is a solution of $\sin v= \dfrac{v}{1964}$. If $v_{0}$ is a solution of $\sin v= \dfrac{v}{1964}$ then $-v_{0}$ is. So we just need to determine the number of positive solutions of $\sin v= \dfrac{v}{1964}$. We have $$v= 1964. \sin v\leqq 1964$$ Therefore, the maximium solution of $\sin v= \dfrac{v}{1964}$ is not over $1964$. We devide ${\rm Ox}$ (also is ${\rm Ov}$) into $312$ ranges of $2\,\pi$ length and a range of $1964- 312.\,2\,\pi> \pi$ length. Based on the nature of 2 function graphs $y= \sin x= \sin v$ and $y= \dfrac{x}{1964}= \dfrac{v}{1964}$ in $(0, 1964)$, we can determine the number of solutions of $\sin v= \dfrac{v}{1964}$ as follows

  • The first range $(0, 2\,\pi)$, then $\sin v= \dfrac{v}{1964}$ has only a positive solution.

  • All the next ranges (except the final range), $\sin v= \dfrac{v}{1964}$ has two positive solutions.

  • The final range $(312.\,2\,\pi, 1964)$, because this range has a larger length than $\pi$, so it 'contains' enough the part of the graph $y= \sin v$ on ${\rm OX}$, therefore, $\sin v= \dfrac{v}{1964}$ also has two positive solutions. In conclusion, the number of solutions of $\sin v= \dfrac{v}{1964}$ is $1+ 311.\,2+ 2= 625$ .

  • The equation $\sin v= \log_{100} v$ is harder to me because I have not studied $\log$ yet. The condition is $x> 0$ and we have $\log_{100} v\leqq 1$. I think we should devide ${\rm Ox}$ (also is ${\rm Ov}$) into many ranges as a.

  • $\begingroup$ The no. of solutions to the first problem is $626$ not $625$ in the first quadrant. You will subtract one occurence of $v = 0$ only when you double the number of the solutions to include the negative ones also. So, total number of solutions are $626 \mul 2 - 1 = 1521$. You can even check this on desmos. $\endgroup$ Jun 23, 2019 at 13:22
  • $\begingroup$ I'll notice that, thank you a real lot ! $\endgroup$
    – user680032
    Jun 23, 2019 at 13:41

1 Answer 1


Hint on # 2:

$\log_{100} v =\dfrac{\log(v)}{\log(100)} $ so $\log(v) \le \log(100)$ so $v \le 100 $.

You then have to divide this into intervals.


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