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Question Number 63645 by mathmax by abdo last updated on 06/Jul/19

$$nintegrnaturalprovethat5divide{n}^5-n$$

Commented by Prithwish sen last updated on 06/Jul/19

$$\mathrm{n}=5\mathrm{k}\Rightarrow {\mathrm{n}}^5-\mathrm{n}\mathrm{divided}\mathrm{by}5$$$$\mathrm{n}=5\mathrm{k}+1\Rightarrow {(5\mathrm{k}+1)}^5-(5\mathrm{k}+1)=5\mathrm{m}\mathrm{divided}\mathrm{by}5$$$$\mathrm{n}=5\mathrm{k}+2\Rightarrow {(5\mathrm{k}+2)}^5-(5\mathrm{k}+2)=5\mathrm{m}+30\mathrm{divided}\mathrm{by}5$$$$\mathrm{n}=5\mathrm{k}+3\Rightarrow {(5\mathrm{k}+3)}^5-(5\mathrm{k}+3)=5\mathrm{m}+240\mathrm{divided}\mathrm{by}5$$$$\mathrm{n}=5\mathrm{k}+4\Rightarrow {(5\mathrm{k}+4)}^5-(5\mathrm{k}+4)=5\mathrm{m}+1020\mathrm{divided}\mathrm{by}5$$$$\mathrm{Hence}\mathrm{proved}.$$

Commented by Prithwish sen last updated on 06/Jul/19

$$\mathrm{Another}\mathrm{method}$$$${\mathrm{n}}^5-\mathrm{n}=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}+1)({\mathrm{n}}^2+1)$$$$\mathrm{for}\mathrm{n}=5\mathrm{k},\mathrm{n}\mathrm{is}\mathrm{divisible}\mathrm{by}5$$$$\mathrm{for}\mathrm{n}=5\mathrm{k}+1,(\mathrm{n}-1)\mathrm{is}\mathrm{divisible}\mathrm{by}5$$$$\mathrm{for}\mathrm{n}=5\mathrm{k}+2,({\mathrm{n}}^2+1)\mathrm{is}\mathrm{divosible}\mathrm{by}5$$$$\mathrm{for}\mathrm{n}=5\mathrm{k}+3,({\mathrm{n}}^2+1)\mathrm{is}\mathrm{divisible}\mathrm{by}5$$$$\mathrm{for}\mathrm{n}=5\mathrm{k}+4,(\mathrm{n}+1)\mathrm{is}\mathrm{divisibleby}5$$$$\therefore {\mathbf{n}}^5-\mathbf{n}\mathbf{is}\mathbf{divisible}\mathbf{by}5.\mathbf{H}\mathrm{ence}\mathrm{proved}.$$$$$$$$$$

Commented by mathmax by abdo last updated on 06/Jul/19

$$thankyousir.$$

Commented by Prithwish sen last updated on 07/Jul/19

$$\mathrm{welcome}$$

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