Mathematical and Physical Journal
for High Schools
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Problem B. 4208. (October 2009)

B. 4208. Let n be a positive integer. Determine the first digit following the decimal point in the number


(Suggested by M. Bencze, Brasov)

(4 pont)

Deadline expired on November 10, 2009.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Minden \(\displaystyle k\) pozitív egész számra

\(\displaystyle (k+0,4)^2=k^2+0,8k+0,16\le k^2+0,96k<k^2+k<k^2+k+0,25=(k+0,5)^2,\)

vagyis \(\displaystyle k+0,4<\sqrt{k(k+1)}<k+0,5\). Ezért

\(\displaystyle \sum_{k=1}^nk+0,4n<\sum_{k=1}^n\sqrt{k(k+1)}< \sum_{k=1}^nk+0,5n,\)


\(\displaystyle \frac{n+1}{2}+0,4< \sum_{k=1}^{n}{\frac{\sqrt{k(k+1)}}{n}}\frac{n+1}{2}+0,5.\)

Ha \(\displaystyle n=2m\), akkor a szóban forgó összeg \(\displaystyle m+0,9\) és \(\displaystyle m+1\) közé esik, ha pedig \(\displaystyle n=2m-1\), akkor \(\displaystyle m+0,4\) és \(\displaystyle m+0,5\) közé.

Vagyis páros \(\displaystyle n\) esetén a kérdéses számjegy 9, páratlan \(\displaystyle n\) esetén pedig 4.


91 students sent a solution.
4 points:63 students.
3 points:1 student.
2 points:8 students.
1 point:2 students.
0 point:10 students.
Unfair, not evaluated:7 solutions.

Problems in Mathematics of KöMaL, October 2009