Solution-1

Assume the contrary, that such numbers exist, and denote them by \(x\) and \(x+1\), obtaining the equation \(x(x+1)=27n+11\).

Multiplying both sides by \(4\) and adding \(1\), we obtain

\[\begin{gathered} 4 x^{2}+4 x+1=108 n+45 \\ (2 x+1)^{2}=108 n+45 \end{gathered}\]

The right-hand side of the equation is divisible by \(9\), hence \(2x+1\) is divisible by \(3\). Dividing both sides of the equation by \(9\), we get

\[\left(\frac{2 x+1}{3}\right)^{2}=12 n+5\]

Note that, when dividing the right-hand side of the equation by \(3\), the remainder is \(2\). However, when dividing a square by \(3\), one can obtain only the remainder \(0\) or \(1\). Therefore this equation has no solution and we have a contradiction. Hence, the product of consecutive positive integers \(x(x+1)\) cannot equal \(27n+11\), where \(n\) is an integer.

Atrisinājums-1

Pienemsim pretējo, ka šādi skaitḷi eksistē un apzīmēsim tos attiecīgi ar \(x\) un \(x+1\), iegūstot vienādojumu \(x(x+1)=27n+11\).

Pareizinot abas puses ar 4 un pieskaitot 1, iegūstam, ka

\[\begin{gathered} 4 x^{2}+4 x+1=108 n+45 \\ (2 x+1)^{2}=108 n+45 \end{gathered}\]

Vienādojuma labā puse dalās ar \(9\), tātad \(2 x+1\) dalās ar \(3\). Izdalot abas vienādojuma puses ar \(9\), iegūstam, ka

\[\left(\frac{2 x+1}{3}\right)^{2}=12 n+5\]

Ievērojam, ka, dalot vienādojuma labo pusi ar 3, tiek iegūts atlikums 2. Toties, dalot skaitla kvadrātu ar 3, var iegūt tikai atlikumu 0 vai 1. Tātad šim vienādojumam nav atrisinājuma un esam ieguvuši pretrunu. Līdz ar to nekādu divu secīgu naturālu skait|u reizinājums nav izsakāms formā \(27 n+11\), kur \(n\) - naturāls skaitlis.