Atrisinājums

Ja \(N=\underbrace{666 \ldots}_{n "6"}\), tad

\(N^{2}=\underbrace{66 \ldots 6}_{n "6"} \cdot \underbrace{66 \ldots 6}_{n "6"}=6 \cdot 6 \cdot \underbrace{11 \ldots 1}_{n "1"} \cdot \underbrace{11 \ldots 1}_{n "1"}=4 \cdot 9 \cdot \underbrace{11 \ldots 1}_{n "1"} \cdot \underbrace{11 \ldots 1}_{n "1"}=\underbrace{44 \ldots}_{n "4"} \cdot \underbrace{99 \ldots 9}_{n "9"}=\)

\(=\underbrace{44 \ldots 4}_{n "4"} \cdot(\underbrace{100 \ldots 0}_{n "0"}-1)=\underbrace{44 \ldots}_{n "4"} \underbrace{00 \ldots 0}_{n "0"}-\underbrace{44 \ldots 4}_{n "4"}=\underbrace{44 \ldots}_{n-1 "4"} 43 \underbrace{55 \ldots .56}_{n-1 "5"}.\)

Atrisinājums

Parādām, kā kāpināt \(66\ldots{}66\) kvadrātā

\[6^2=36,\;\;66^2=4356=4455-99,\;\;666^2=443556=444555-999,\ldots\]

Pamatosim, ka

\[\begin{aligned} (\underbrace{6\ldots6}_n)^2 & = \overline{\underbrace{4\ldots{}4}_n\underbrace{5\ldots{}5}_n}-\underbrace{9\ldots{}9}_n, \\ \left( 6\cdot(10^n-1)/9 \right)^2 & = 10^n \cdot (4 \cdot (10^n - 1)/9) +(5 \cdot (10^n - 1)/9) - (10^n-1), \\ \frac{4}{9}(10^n-1)^2 & = 10^n \cdot \frac{4}{9}(10^n - 1) +\frac{5}{9}(10^n-1) - (10^n - 1), \\ 4(10^n - 1)^2 & = 4\cdot{}10^{2n}-4\cdot{}10^n + 5\cdot{}10^n-5 - 9\cdot{}10^n +9. \end{aligned}\]

Tātad \(\overline{6\ldots6}^2\) pierakstā ir tikai cipari "4","3", "5" un "6":

\[(\underbrace{6\ldots6}_n)^2=\overline{\underbrace{4\ldots{}4}\_{n-1}3\underbrace{5\ldots{}5}\_{n-1}6}\]