S:E15.208

S:E15.208 (Angela Lopată)

Determinați toate numerele naturale consecutive care au suma Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2015} .

Soluția 1 Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\in \mathbb{N}} și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N\in \mathbb{N}\setminus\left\{0,1\right\}} numere naturale pentru care Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(a+1\right) + \left(a+2\right)+\ldots+\left(a+N\right)=2015.} În mod echivalent, se obține $Na+\left(1+2+\ldots +N\right)=2015$, deci

$$N\cdot \left(2a+1+N\right)=2\cdot 2015.$$

Din Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\ge 0} , avem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N\left(N+1\right)\le N\left(2a+1+N\right) = 2024} . Cum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 63\cdot 64 \le 4030 \le 64\cdot 65} , se deduce c\u a Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N\le 63} .

Din <math>N | 4030<math> \c si <math>N\le 63<math> rezult\u a <math>N\in \left\{2,5,10,13,26,31, 62 \right\}<math>. \\ Pentru <math>N=2<math> se ob\c tine <math>2a+3 = 2015<math>, cu <math>a =1006<math>. Deci avem suma de două umere consecutive <math display="block"> \label{eq2cls6} 1007+1008=2015. </math display="block">

Pentru <math>N=5<math> se ob\c tine <math>2a+6 = 806<math>, cu <math>a = 400<math>. Deci avem suma de <math>5<math> numere consecutive

<math display="block"> \label{eq3cls6} 401+402+403+404+405=2015. </math display="block"> Pentru <math>N=10<math> se ob\c tine <math>2a+11 = 403<math>, cu <math>a = 196<math>. Deci avem suma de <math>10<math> numere consecutive <math display="block"> \label{eq4cls6} 197+198+\ldots+206=2015. </math display="block"> Pentru <math>N=13<math> se ob\c tine <math>2a+14 = 310<math>, cu <math>a = 148<math>. Deci avem suma cu <math>13<math> termeni, numere consecutive <math display="block"> \label{eq5cls6} 149+150+\ldots+161=2015. </math display="block"> Pentru <math>N=26<math> se ob\c tine <math>2a+27 = 155<math>, cu <math>a = 64<math>. Deci avem suma de <math>26<math> de numere consecutive <math display="block"> \label{eq6cls6} 65+66+\ldots+67=2015. </math display="block"> Pentru <math>N=31<math> se ob\c tine <math>2a+32 = 130<math>, cu <math>a = 49<math>. Deci avem suma de <math>31<math> de numere consecutive <math display="block"> \label{eq8cls6} 50+51+\ldots+80=2015. </math display="block"> Pentru <math>N=62<math> se ob\c tine <math>2a+63 = 65<math>, cu <math>a = 1<math>. Deci avem suma de <math>62 <math> de numere consecutive <math display="block"> \label{eq7cls6} 2+3+\ldots+63=2015. </math display="block">

Soluția 2