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

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.

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