Mathematical and Physical Journal
for High Schools
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Problem K. 403. (January 2014)

K. 403. How many digits of 8 are there in the decimal form of the sum 0.9+0.99+0.999+\dots+0.
\underbrace{999\dots 9}_{2013\; \rm{pieces}}?

(6 pont)

Deadline expired on February 10, 2014.


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

Megoldás. Írjuk az összeget más alakban!

0{,}9+0{,}99+0{,}999+\dots+0{,}
\underbrace{999\dots 9}_{2013\;\rm{db}}=(1-0,1)+(1-0,01)+(1-0,001)+
...+(1-0{,}
\underbrace{000\dots 01}_{2013\;\rm{db}}) =

= 2013-0{,}
\underbrace{111\dots 1}_{2013\;\rm{db}}=2012{,}
\underbrace{888\dots 89}_{2013\;\rm{db}}.

Tehát az összeg tizedestört alakja 2012 darab 8-as számjegyet tartalmaz.


Statistics:

180 students sent a solution.
6 points:106 students.
5 points:36 students.
4 points:11 students.
3 points:12 students.
2 points:2 students.
1 point:4 students.
Unfair, not evaluated:9 solutionss.

Problems in Mathematics of KöMaL, January 2014