{"batchcomplete":"","continue":{"lecontinue":"20250916033521|3301","continue":"-||"},"query":{"logevents":[{"logid":3311,"ns":0,"title":"E:16892","pageid":3104,"logpage":3104,"revid":10785,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-20T13:32:32Z","comment":"Created page with \"'''[[E:16892]] (Nicolae Mu\u0219uroia)'''  ''Afla\u021bi suma divizorilor pari ai celui mai mare num\u0103r natural <math>a</math>, cu <math>a<1000</math>, pentru  care suma divizorilor impari este egal\u0103 cu <math>24</math>.''  '''Solu\u021bie'''  C\u0103ut\u0103m numere de trei cifre, de forma <math>2^m \\cdot b</math>, cu <math>m\\in \\mathbb{N}^\\ast</math> \u0219i <math>b</math>, unde suma divizorilor num\u0103rului natural impar <math>b</math> este egal\u0103 cu <math>24</math>.  Avem dou\u0103 posibilit\u0103\u021bi...\""},{"logid":3310,"ns":0,"title":"E:16893","pageid":3103,"logpage":3103,"revid":10781,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-20T13:12:15Z","comment":"Created page with \"'''[[E:16893]] (Traian Covaciu)'''  ''Ar\u0103ta\u021bi c\u0103 numerele <math>7n-1</math> \u0219i <math>17n-1</math> sunt simultan prime doar dac\u0103 <math>n</math> este un multiplu natural al lui <math>6</math>.''  '''Solu\u021bie'''  Pentru <math>n=6</math> se ob\u021bin numerele prime <math>42</math> \u0219i <math>101</math>.  Dac\u0103 <math>n</math> este impar, atunci numerele <math>7n-1</math> \u0219i <math>17n-1</math> sunt pare, deci nu pot fi prime, ceea ce implic\u0103 faptul c\u0103 <math>2 |\\, n</math>....\""},{"logid":3309,"ns":0,"title":"E:16891","pageid":3102,"logpage":3102,"revid":10778,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-20T12:58:44Z","comment":"Created page with \"'''[[E:16891]] (Sever Pop)'''  ''Determina\u021bi numerele prime <math>p</math>, <math>q</math>, <math>r</math>, distincte dou\u0103 c\u00e2te dou\u0103, pentru care are loc egalitatea <math>3p^4 - 5q^4 - 4r^2 = 26</math>.''  '''Solu\u021bie'''  ''Deoarece <math>3p^4 - 5q^4=2\\left(13+2r^2\\right)</math> este num\u0103r par, deducem c\u0103 numerele prime <math>p</math> \u0219i <math>q</math> au aceea\u0219i paritate, deci sunt impare.  Cum <math>3p^4 - 5q^4>26>0</math>, avem <math>3p^3>5q^4</math>, deci <ma...\""},{"logid":3308,"ns":0,"title":"E:16889","pageid":3101,"logpage":3101,"revid":10775,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-20T05:42:38Z","comment":"Created page with \"'''[[E:16889]] (C\u0103lin Hossu)'''  ''Prin \u00eemp\u0103r\u021birea unui num\u0103r de patru cifre la r\u0103sturnatul s\u0103u, se ob\u021bine c\u00e2tul <math>2</math> \u0219i restul <math>1977</math>. Afla\u021bi num\u0103rul, \u0219tiind c\u0103 diferen\u021ba dintre cifra miilor \u0219i cifra unit\u0103\u021bilor este <math>5</math>, iar cifra sutelor este cu <math>4</math> mai mare dec\u00e2t cifra zecilor.''  '''Solu\u021bie'''\""},{"logid":3307,"ns":0,"title":"E:16890","pageid":3100,"logpage":3100,"revid":10772,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-20T05:36:12Z","comment":"Created page with \"'''[[E:16890]] (Bogdan Zetea, C\u0103lin Hossu)'''  ''Demonstra\u021bi c\u0103, pentru orice num\u0103r natural nenul <math>n</math>, num\u0103rul <math>2024^n+n^{2024} + 2</math> nu este un p\u0103trat perfect.''  '''Solu\u021bie'''  Fie <math>N = 2024^n+n^{2024} + 2</math>.  Dac\u0103 <math>n</math> este un num\u0103r par, atunci exist\u0103 numerele naturale nenule <math>t</math> \u0219i <math>u</math> pentru care <math>n^{2024} = 4t</math> \u0219i <math>2024^n = 4u</math>. Atunci <math>N = 4t+4u+2 <math>, deci exi...\""},{"logid":3306,"ns":0,"title":"E:16888","pageid":3099,"logpage":3099,"revid":10767,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-19T19:32:21Z","comment":"Created page with \"'''[[E:16888]] (Gheorghe Boroica)'''  ''Consider\u0103m <math>n<math> un num\u0103r natural nenul. Demonstra\u021bi c\u0103 num\u0103rul <math>N = \\underbrace{44\\ldots4}_{n \\text{  cifre}}\\underbrace{22\\ldots2}_{n \\text{ cifre}} </math>  poate fi scris ca produsul a dou\u0103 numere naturale consecutive.''  ''''Solu\u021bie'''  Dac\u0103 <math>a=\\underbrace{11\\ldots1}_{n \\text{  cifre}}</math>, atunci <math>9\\cdot a+1=10^n</math> \u0219i <math display=\"block\">N= 4\\cdot a \\cdot 10^n + 2 \\cdot a = 2\\cdot a \\...\""},{"logid":3305,"ns":0,"title":"E:16887","pageid":3098,"logpage":3098,"revid":10765,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-19T19:29:49Z","comment":"Created page with \"'''[[E:16887]] (Gheorghe Boroica)'''  ''Suma a <math>90</math> de numere naturale este <math>2069</math>. Ar\u0103ta\u021bi c\u0103 exist\u0103, printre acestea, cel pu\u021bin trei numere egale.''  '''Solu\u021bie'''  Fie <math>S</math> suma celor <math>90</math> de numere. Presupunem contrariul, deci printre cele <math>90</math> de numere, cel mult dou\u0103 numere pot fi egale. Atunci  <math display=\"block\"> \tS  \\ge \\left(1+1\\right) + \\left(2+2\\right) + \\left(3+3\\right)+\\ldots +\\left(45+45\\right...\""},{"logid":3304,"ns":0,"title":"P:1800","pageid":3097,"logpage":3097,"revid":10761,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-19T16:57:10Z","comment":"Created page with \"'''[[P:1800]] (Ioan Ovidiu Pop, Coroieni)'''  ''Afla\u021bi num\u0103rul de telefon <math>\\overline{07abcdefgh}</math>, format din zece cifre, nu neap\u0103rat distincte, pentru care numerele <math>a+c</math>, <math>b+c</math>, <math>d+e</math>, <math>c+d</math>, <math>a+b+e</math>, <math>c+d+f</math>, <math>b+c+g</math> \u0219i <math>d+e+h</math> sunt opt numere consecutive a\u0219ezate \u00een ordine cresc\u0103toare.''  '''Solu\u021bie''' Cele opt numere consecutive a\u0219ezate \u00een ordine cresc\u0103toare...\""},{"logid":3303,"ns":0,"title":"23964","pageid":3096,"logpage":3096,"revid":10758,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-17T17:50:42Z","comment":"Created page with \"'''[[23964]] (Marin Banco\u0219)'''  ''S\u0103 de demonstreze inegalitatea <math display=\"block\"> \\sum_{i=2}^{n} \\sqrt[i]{\\left(i!\\right)^2} < \\frac{2n^3+9n^2+13n-24}{24} .</math>''  '''Solu\u021bie''' Pentru orice num\u0103r natural <math>n</math>, cu <math> n\\ge 2</math> are loc inegalitatea <math display=\"block\"> \\sqrt[n]{n!} = \\sqrt[n]{1\\cdot 2\\cdot \\ldots \\cdot n} < \\frac{1+2+\\ldots+n}{n} = \\frac{n+1}{2}.</math> Atunci  <math display=\"block\"> \\sum_{i=2}^{n} \\sqrt[i]{\\left(I!\\right)...\""},{"logid":3302,"ns":0,"title":"Gazeta matematic\u0103 1998","pageid":3095,"logpage":3095,"revid":10754,"params":{},"type":"create","action":"create","user":"Andrei.Horvat","timestamp":"2025-09-17T17:48:23Z","comment":"Created page with \"== Gazeta Matematic\u0103 9/1998 ==\""}]}}