Bruker:Karl Erik/Problem10 - Sideversjonshistorikk

https://matematikk.net/w/index.php?action=history&feed=atom&title=Bruker%3AKarl_Erik%2FProblem10 Bruker:Karl Erik/Problem10 - Sideversjonshistorikk 2026-04-09T04:13:31Z Versjonshistorikk for denne siden på wikien MediaWiki 1.42.3 https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem10&diff=3649&oldid=prev Jarle: Tømmer siden 2011-02-04T18:48:36Z

<p>Tømmer siden</p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 18:48</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1">Linje 1:</td> <td colspan="2" class="diff-lineno">Linje 1:</td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Problem10</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">----</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Define <tex>f(n) = \sum^n_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{n}</tex>. We will be using that <tex>\sum^m_{k=1}\cos \frac{2 \pi k}{m} = \Re(\sum^m_{k=1}e^{i\frac{2 \pi k}{m}}) = \Re(e^{i\frac{2 \pi}{m}}\frac{e^{mi\frac{2 \pi}{m}}-1}{e^{i\frac{2 \pi}{m}-1}}) = 0</tex>.</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Consider a prime p. Then <tex>f(p) = \sum^p_{k=1} \gcd(k,p)\cos \frac{2 \pi k}{p} = \sum^p_{k=1} \cos \frac{2 \pi k}{p} +(p-1)\cos \frac{2 \pi p}{p} = p-1</tex>. We will show that if <tex>a_1,a_2,...,a_r</tex> are distinct prime factors, then <tex>f(a_1...a_r) = (a_1-1)...(a_r-1)</tex> by induction on <tex>r</tex>. We have already shown this for <tex>r = 1</tex>, so suppose it is true for <tex>s \leq r</tex>. </del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Let <tex>n = a_1...a_s</tex>, and let <tex>a_{s+1}</tex> be a prime not among the prime factors of <tex>n</tex>. <tex>f(na_{s+1}) = \sum^{na_{s+1}}_{k=1} \gcd(k,na_{s+1})\cos \frac{2 \pi k}{na_{s+1}} = \sum^{na_{s+1}}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{na_{s+1}} + (a_{s+1}-1)\sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{n} \\ =\sum^{na_{s+1}}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{na_{s+1}} +(a_{s+1}-1)f(n)</tex>. </del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">So it remains to show that <tex>\sum^{np}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} = 0</tex> whenever <tex>p</tex> is a prime not among the distinct prime factors of <tex>n</tex>. Now, <tex>\sum^{np}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} = </del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">\sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} + \sum^{2n}_{k=n+1} \gcd(k,n)\cos \frac{2 \pi k}{np} + ... + \sum^{np}_{k=n(p-1)+1} \gcd(k,n)\cos \frac{2 \pi k}{np} \\ = \sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} + \sum^{n}_{k=1} \gcd(k+n,n)\cos \frac{2 \pi (k+n)}{np} + ... + \sum^{n}_{k=1} \gcd(k+(p-1)n,n)\cos \frac{2 \pi (k+(p-1)n)}{np} \\ = \sum^{n}_{k=1} \gcd(k,n)\left(\cos \frac{2 \pi k}{np} +\cos (\frac{2 \pi k}{np}+\frac{2\pi}{p}) + \cos (\frac{2 \pi k}{np}+\frac{2\pi2}{p}) + ... + \cos (\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p}) \right)</tex>.</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">But for each <tex>k</tex> between <tex>1</tex> and <tex>n</tex>, we have <tex>\cos \frac{2 \pi k}{np} +\cos (\frac{2 \pi k}{np}+\frac{2\pi}{p}) + \cos (\frac{2 \pi k}{np}+\frac{2\pi2}{p}) + ... + \cos (\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p}) = \Re(e^{ i\frac{2 \pi k}{np}} +e^{i (\frac{2 \pi k}{np}+\frac{2\pi}{p})} + e^{i (\frac{2 \pi k}{np}+\frac{2\pi2}{p})} + ... + e^{i(\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p})}) \\ = \Re(e^{ i\frac{2 \pi k}{np}}(1+e^{i \frac{2\pi}{p}} + e^{i\frac{2\pi2}{p}} + ... + e^{i\frac{2\pi(p-1)}{p})}) = 0</tex>.</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Thus we conclude that <tex>f(na_{s+1})=(a_{s+1}-1)f(n)=(a_1-1)...(a_s-1)(a_{s+1}-1)</tex>, and we are done.</del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2" class="diff-side-added"></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Now, <tex>2010 = 2 \times 3 \times 5 \times 67</tex> is a product of distinct prime factors, so <tex>\sum^{2010}_{k=1} \gcd(k,2010)\cos \frac{2 \pi k}{2010} = f(2010) = (2-1)(3-1)(5-1)(67-1)=528</tex>.</del></div></td><td colspan="2" class="diff-side-added"></td></tr>  </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem10&diff=3618&oldid=prev Jarle på 4. feb. 2011 kl. 02:29 2011-02-04T02:29:11Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="nb"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Eldre sideversjon</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Sideversjonen fra 4. feb. 2011 kl. 02:29</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1">Linje 1:</td> <td colspan="2" class="diff-lineno">Linje 1:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Problem10</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Problem10</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>----</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>----</div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Define <tex>f(n) = \sum^n_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{n}</tex>. We will be using that <tex>\sum^m_{k=1}\cos \frac{2 \pi k}{m} = \Re(\sum^m_{k=1}e^{i\frac{2 \pi k}{m}}) = \Re(e^{i\frac{2 \pi}{m}}\frac{e^{mi\frac{2 \pi}{m}}-1}{e^{i\frac{2 \pi}{m}-1}}) = 0</tex>.</ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Consider a prime p. Then <tex>f(p) = \sum^p_{k=1} \gcd(k,p)\cos \frac{2 \pi k}{p} = \sum^p_{k=1} \cos \frac{2 \pi k}{p} +(p-1)\cos \frac{2 \pi p}{p} = p-1</tex>. We will show that if <tex>a_1,a_2,...,a_r</tex> are distinct prime factors, then <tex>f(a_1...a_r) = (a_1-1)...(a_r-1)</tex> by induction on <tex>r</tex>. We have already shown this for <tex>r = 1</tex>, so suppose it is true for <tex>s \leq r</tex>. </ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Let <tex>n = a_1...a_s</tex>, and let <tex>a_{s+1}</tex> be a prime not among the prime factors of <tex>n</tex>. <tex>f(na_{s+1}) = \sum^{na_{s+1}}_{k=1} \gcd(k,na_{s+1})\cos \frac{2 \pi k}{na_{s+1}} = \sum^{na_{s+1}}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{na_{s+1}} + (a_{s+1}-1)\sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{n} \\ =\sum^{na_{s+1}}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{na_{s+1}} +(a_{s+1}-1)f(n)</tex>. </ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">So it remains to show that <tex>\sum^{np}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} = 0</tex> whenever <tex>p</tex> is a prime not among the distinct prime factors of <tex>n</tex>. Now, <tex>\sum^{np}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} = </ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} + \sum^{2n}_{k=n+1} \gcd(k,n)\cos \frac{2 \pi k}{np} + ... + \sum^{np}_{k=n(p-1)+1} \gcd(k,n)\cos \frac{2 \pi k}{np} \\ = \sum^{n}_{k=1} \gcd(k,n)\cos \frac{2 \pi k}{np} + \sum^{n}_{k=1} \gcd(k+n,n)\cos \frac{2 \pi (k+n)}{np} + ... + \sum^{n}_{k=1} \gcd(k+(p-1)n,n)\cos \frac{2 \pi (k+(p-1)n)}{np} \\ = \sum^{n}_{k=1} \gcd(k,n)\left(\cos \frac{2 \pi k}{np} +\cos (\frac{2 \pi k}{np}+\frac{2\pi}{p}) + \cos (\frac{2 \pi k}{np}+\frac{2\pi2}{p}) + ... + \cos (\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p}) \right)</tex>.</ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">But for each <tex>k</tex> between <tex>1</tex> and <tex>n</tex>, we have <tex>\cos \frac{2 \pi k}{np} +\cos (\frac{2 \pi k}{np}+\frac{2\pi}{p}) + \cos (\frac{2 \pi k}{np}+\frac{2\pi2}{p}) + ... + \cos (\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p}) = \Re(e^{ i\frac{2 \pi k}{np}} +e^{i (\frac{2 \pi k}{np}+\frac{2\pi}{p})} + e^{i (\frac{2 \pi k}{np}+\frac{2\pi2}{p})} + ... + e^{i(\frac{2 \pi k}{np}+\frac{2\pi(p-1)}{p})}) \\ = \Re(e^{ i\frac{2 \pi k}{np}}(1+e^{i \frac{2\pi}{p}} + e^{i\frac{2\pi2}{p}} + ... + e^{i\frac{2\pi(p-1)}{p})}) = 0</tex>.</ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Thus we conclude that <tex>f(na_{s+1})=(a_{s+1}-1)f(n)=(a_1-1)...(a_s-1)(a_{s+1}-1)</tex>, and we are done.</ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Now, <tex>2010 = 2 \times 3 \times 5 \times 67</tex> is a product of distinct prime factors, so <tex>\sum^{2010}_{k=1} \gcd(k,2010)\cos \frac{2 \pi k}{2010} = f(2010) = (2-1)(3-1)(5-1)(67-1)=528</tex>.</ins></div></td></tr>  </table>

Jarle https://matematikk.net/w/index.php?title=Bruker:Karl_Erik/Problem10&diff=3610&oldid=prev Karl Erik: Ny side: Problem10 ---- 2011-02-03T22:18:53Z

<p>Ny side: Problem10 ----</p> <p><b>Ny side</b></p><div>Problem10<br /> ----</div>

Karl Erik