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Concursul "Congruente", problema 1
Posted: Sun Nov 09, 2008 10:55 pm
by Claudiu Mindrila
Cate numere cuprinse intre \( 100 \) si \( 801 \) se impart exact la \( 8 \)?
Gazeta Matematica
Posted: Mon Nov 10, 2008 10:58 pm
by naruto
\( 100 < 8k < 801 \)
\( 100=8\cdot 12+4 \)
\( 801=8\cdot 100+1
\)
Rezulta \( 8k \in \{8\cdot 13, 8\cdot 14,........,8\cdot 100\} \).
\( 100-13+1=88 \). Raspuns: \( \overline{ \underline{|\ 88 \ \mbox numere\ |} } \)
Posted: Wed Nov 19, 2008 11:15 am
by Dorobantu Razvan
intre 100 si 801 sunt cuprinse 700 nr.
primul care se imparte la 8 este 104, iar ultimul 800
intre acestea sunt cuprinse 697 nr. (inclusiv 104 si 800)
697:8=87(rest 1)
87+1=88 nr.
E bine?

Posted: Wed Nov 19, 2008 7:49 pm
by Marcelina Popa
Da, e corect

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