Chứng minh rằng $5^{n+2}+26.5^{n}+8^{2n+1}$ $\vdots$ $59$
Chứng minh rằng $5^{n+2}+26.5^{n}+8^{2n+1}$ $\vdots$ $59$
Started By grigoriperelmanlapdi, 02-07-2015 - 18:58
#2
Posted 02-07-2015 - 20:10
LeHKhai
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Chứng minh rằng $5^{n+2}+26.5^{n}+8^{2n+1}$ $\vdots$ $59$
$5^{n+2}+26.5^n+8^{2n+1}=5^{n}.25+26.5^{n}+8.8^{2n}=51.5^{n}+8.64^{n}=8\left ( 64^{n}-5^{n} \right )+59.5^{n}=8\left ( 64-5 \right )\left ( ... \right )+59.5^{n}\vdots 59$
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#3
Posted 04-07-2015 - 10:29
Hoangtheson2611
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$5^{n+2}+26.5^{n}+8^{2n+1}\equiv 5^{n}.25+5^{n}.26+5^{n}.8\equiv 5^{n}.59 (mod 59)$
=>$5^{n+2}+26.5^{n}+8^{2n+1}$ chia hết cho 59
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