Prezint o generalizare a problemei 4 de la TMMATE 2008
Daca notam cu \( p(n) \) produsul cifrelor nenule ale lui \( n \) atunci demonstrati ca \( \sum_{i=1}^{10^k}p(i) \) este puterea \( k \) a unui numar natural.
La concurs problema a fost data pentru \( k=3 \).
Generalizare produs cifre
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Generalizare produs cifre
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Demonstram ca \( \sum_{i=1}^{10^k}p(i) = 46^k \)
Inductie: pentru \( k=1 \) evident \( \sum_{i=1}^{10} =46 \)
Pentru \( k=2 \) \( \sum_{i=1}^{100} =46+46+46*2+46*3+...+46*9=46(1+1+2+...+9)=
46*46= 46^2 \)
Presupunem ca pentru \( k=n \) propozitia "\( P(k): \sum_{i=1}^{10^k} = 46^k \) " e adevarata si demonstram propozitia pentru \( k=n+1 \).
\( \sum_{i=1}^{10^{(n+1)}} =46^n+46^n + 2*46^n+3*46^n+...+9*46^n=
(1+1+2+3+...+9)* 46^n=46*46^n=46^{(n+1)} \).
Deci suma data e puterea k a numarului 46,care e natural.
Inductie: pentru \( k=1 \) evident \( \sum_{i=1}^{10} =46 \)
Pentru \( k=2 \) \( \sum_{i=1}^{100} =46+46+46*2+46*3+...+46*9=46(1+1+2+...+9)=
46*46= 46^2 \)
Presupunem ca pentru \( k=n \) propozitia "\( P(k): \sum_{i=1}^{10^k} = 46^k \) " e adevarata si demonstram propozitia pentru \( k=n+1 \).
\( \sum_{i=1}^{10^{(n+1)}} =46^n+46^n + 2*46^n+3*46^n+...+9*46^n=
(1+1+2+3+...+9)* 46^n=46*46^n=46^{(n+1)} \).
Deci suma data e puterea k a numarului 46,care e natural.
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