Mathematical and Physical Journal
for High Schools
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Problem K. 64. (December 2005)

K. 64. Express the following sum as a ratio of two integers:


\frac{4}{1989\cdot 1993}+ \frac{4}{1993\cdot 1997}+ \frac{4}{1997\cdot 2001}+
\frac{4}{2001\cdot 2005}.

(6 pont)

Deadline expired on January 10, 2006.


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

Megoldás: Az első törtet felírhatjuk két tört különbségeként: {4\over1989\cdot1993}={1\over1989}-{1\over1993}. Hasonlóan felírva a többit is, a négytagú összeget így írhatjuk más alakban:

{1\over1989}-{1\over1993}+{1\over1993}-{1\over1997}+{1\over1997}-{1\over2001}+{1\over2001}-{1\over2005}=

{1\over1989}-{1\over2005}={2005-1989\over1989\cdot2005}={16\over3987945}.


Statistics:

176 students sent a solution.
6 points:86 students.
5 points:19 students.
4 points:29 students.
3 points:23 students.
2 points:3 students.
1 point:4 students.
0 point:12 students.

Problems in Mathematics of KöMaL, December 2005