Consideram \( n \in \mathbb{N} \), \( n \ge 4 \). Aratati ca daca \( \left\[ \frac{2^n}{n} \right\] \) este o putere a lui \( 2 \) atunci \( n \) este o putere a lui \( 2 \).
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Filip Chindea
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Parte intreaga si puteri
Life is complex: it has real and imaginary components.
Fie \( n=2^r(2s+1) \) cu \( r,s\in N \) a.i. \( n\geq4 \).
\( \left[\frac{2^n}{n}\right]=\left[\frac{2^{2^r(2s+1)}}{2^r(2s+1)}\right]=\left[\frac{2^{2^r(2s+1)-r}}{2s+1}\right]=2^k \), pt un \( k\leq2^r(2s+1)-r=\alpha \).
Deci \( 2^{\alpha}=2^k(2s+1)+p \), cu \( 0\leq p<2s+1 \). Asta implica \( p=a*2^k \) cu \( 0\leq a <\frac{2s+1}{2^k},a\in N \).
Demonstrez ca \( a=0 \).
Fie \( f(x)=\left[\frac{2^x}{x}\right]\quad f:N-\left{0,1,2,3\right}\rightarrow N \). Se demonstreaza usor inductiv ca \( f(x)\geq x \), \( \forall x \). Si mai avem si ca \( \left[\frac{2^y}{x}\right]\geq\left[\frac{2^x}{x}\right] \) cand \( y\geq x \).
Cum \( 2^r(2s+1)-r\geq 2s+1 \) din cauza ca \( 2^r\geq r+1 \) din Bernoulli, luam \( y=\alpha \) si \( x=2s+1 \). Astfel
\( 2^k=\left[\frac{2^{\alpha}}{2s+1}\right]\geq \left[\frac{2^{2s+1}}{2s+1}\right]\geq 2s+1 \). Atunci \( 0\leq a < \frac{2s+1}{2^k}\leq 1 \) implica \( a=0 \) ceea ce implica \( 2s+1=2^{\alpha-k}\rightarrow s=0\rightarrow n=2^r \), cu \( r\in N-\left{0,1\right} \).
Asta a fost solutia mea din concurs.
\( \left[\frac{2^n}{n}\right]=\left[\frac{2^{2^r(2s+1)}}{2^r(2s+1)}\right]=\left[\frac{2^{2^r(2s+1)-r}}{2s+1}\right]=2^k \), pt un \( k\leq2^r(2s+1)-r=\alpha \).
Deci \( 2^{\alpha}=2^k(2s+1)+p \), cu \( 0\leq p<2s+1 \). Asta implica \( p=a*2^k \) cu \( 0\leq a <\frac{2s+1}{2^k},a\in N \).
Demonstrez ca \( a=0 \).
Fie \( f(x)=\left[\frac{2^x}{x}\right]\quad f:N-\left{0,1,2,3\right}\rightarrow N \). Se demonstreaza usor inductiv ca \( f(x)\geq x \), \( \forall x \). Si mai avem si ca \( \left[\frac{2^y}{x}\right]\geq\left[\frac{2^x}{x}\right] \) cand \( y\geq x \).
Cum \( 2^r(2s+1)-r\geq 2s+1 \) din cauza ca \( 2^r\geq r+1 \) din Bernoulli, luam \( y=\alpha \) si \( x=2s+1 \). Astfel
\( 2^k=\left[\frac{2^{\alpha}}{2s+1}\right]\geq \left[\frac{2^{2s+1}}{2s+1}\right]\geq 2s+1 \). Atunci \( 0\leq a < \frac{2s+1}{2^k}\leq 1 \) implica \( a=0 \) ceea ce implica \( 2s+1=2^{\alpha-k}\rightarrow s=0\rightarrow n=2^r \), cu \( r\in N-\left{0,1\right} \).
Asta a fost solutia mea din concurs.
Iata si o schita a unei alte solutii.
Fie \( 2^{k}=\[\frac{2^{n}}{n}\]\Leftrightarrow 2^{k}\leq \frac{2^{n}}{n}<2^{k}+1\Leftrightarrow 2^{k}n\leq 2^{n} < 2^{k}n+n\Leftrightarrow n\leq 2^{n-k} < n+\frac{n}{2^{k}}. \) Dar \( n\leq 2^{k} \), pentru \( n\geq 5. \) Intr-adevar, in caz contrar, daca \( n>2^{k}\Rightarrow n\geq\frac{2^{n}}{n}, \) adica \( n^{2}\geq 2^{n}. \) Dar se verifica usor prin inductie dupa \( n\geq 5 \) ca \( n^{2}<2^{n} \), deci \( n\leq 2^{n-k} < n+\frac{n}{2^{k}}\leq n+1, \) deci \( n=2^{n-k}, \) \( k\leq n. \) \( n=4 \) de asemenea verifica
iar concluzia se impune.
Fie \( 2^{k}=\[\frac{2^{n}}{n}\]\Leftrightarrow 2^{k}\leq \frac{2^{n}}{n}<2^{k}+1\Leftrightarrow 2^{k}n\leq 2^{n} < 2^{k}n+n\Leftrightarrow n\leq 2^{n-k} < n+\frac{n}{2^{k}}. \) Dar \( n\leq 2^{k} \), pentru \( n\geq 5. \) Intr-adevar, in caz contrar, daca \( n>2^{k}\Rightarrow n\geq\frac{2^{n}}{n}, \) adica \( n^{2}\geq 2^{n}. \) Dar se verifica usor prin inductie dupa \( n\geq 5 \) ca \( n^{2}<2^{n} \), deci \( n\leq 2^{n-k} < n+\frac{n}{2^{k}}\leq n+1, \) deci \( n=2^{n-k}, \) \( k\leq n. \) \( n=4 \) de asemenea verifica
iar concluzia se impune.