Question

Arată că:
a)
${2}^{3} \times {5}^{2} - {2}^{2} \times 5$
este multiplu de 9

b)
${2}^{4} \times {3}^{2} \times {5}^{3} + {5}^{5}$
este multiplu de 13

c)
${10}^{n} + 2$
este multiplu de 3





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Answer 1

 

$\displaystyle\bf\\a)\\\\2^3\times5^2-2^2\times5=\\\\=2^{2+1}\times5^{1+1}-2^2\times5=\\\\=2^2\times2\times5\times5-2^2\times5=\\\\=2^2\times5\times2\times5-2^2\times5=~~~~ Dam~factor~comun~pe~2^2\times5\\\\=2^2\times5\big(2\times5-1\big)=\\\\=2^2\times5\big(10-1\big)=\\\\=2^2\times5\times9~\vdots~9~~deoarece~un~factor~=9$

.

$\displaystyle\bf\\b)\\\\2^4\times3^2\times5^3+5^5=\\\\=5^3\Big(2^4\cdot3^2+5^2\Big)=\\\\=5^3\Big(16\cdot9+5^2\Big)=\\\\=5^3\Big(144+5^2\Big)=\\\\=5^3\Big(12^2+5^2\Big)=\\\\~~~~~~~~~~~~~~~~~~~~~~~~12^2+5^2=13^2\\~~~~~~~~~~~~~~~~~~~~~~~~deoarece~5;~12~si~13~sunt~numere~pitagorice\\\\ =5^3\times13^2~\vdots~13$

.

$\fdisplaystyle\bf\\c)\\\\10^n+2\\\\Cazul~1:\\n=0\\\\10^n+2=10^0+2=1+2=3~\vdots~3\\\\Cazul~2:\\n=1\\\\10^n+2=10^1+2=10+2=12~\vdots~3\\\\Cazul~2:\\n\geq2\\\\10^n+2=1\!\underbrace{\bf00000...00}_{n~zerouri}+2=1\!\underbrace{\bf00000...000}_{(n-1)~zerouri}\!2\\\\Soma~cifrelor~este~1+\underbrace{0+0+0+0+...+0+0}_{(n-1)~zerouri}+2=3\\\\\implies~(10^n+2)~\vdots~3~pentru~\forall~n\in N$