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Question Number 33462 by NECx last updated on 17/Apr/18
using PMI show that ∀ n≥2 the  number 5^n  ends with the digits 25
$${using}\:{PMI}\:{show}\:{that}\:\forall\:{n}\geqslant\mathrm{2}\:{the} \\ $$$${number}\:\mathrm{5}^{{n}} \:{ends}\:{with}\:{the}\:{digits}\:\mathrm{25} \\ $$
Answered by MJS last updated on 17/Apr/18
base:  n=2  5^2 =25    if true for n ⇒ true for n+1  i, j ∈ N  5^n =100i+25 ⇒ 5^(n+1) =100j+25  5^(n+1) =(100i+25)×5=500i+125=  =500i+100+25=100j+25 with  j=5i+1  this holds for n≥2
$${base}: \\ $$$${n}=\mathrm{2} \\ $$$$\mathrm{5}^{\mathrm{2}} =\mathrm{25} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{true}\:\mathrm{for}\:{n}\:\Rightarrow\:\mathrm{true}\:\mathrm{for}\:{n}+\mathrm{1} \\ $$$${i},\:{j}\:\in\:\mathbb{N} \\ $$$$\mathrm{5}^{{n}} =\mathrm{100}{i}+\mathrm{25}\:\Rightarrow\:\mathrm{5}^{{n}+\mathrm{1}} =\mathrm{100}{j}+\mathrm{25} \\ $$$$\mathrm{5}^{{n}+\mathrm{1}} =\left(\mathrm{100}{i}+\mathrm{25}\right)×\mathrm{5}=\mathrm{500}{i}+\mathrm{125}= \\ $$$$=\mathrm{500}{i}+\mathrm{100}+\mathrm{25}=\mathrm{100}{j}+\mathrm{25}\:\mathrm{with} \\ $$$${j}=\mathrm{5}{i}+\mathrm{1} \\ $$$$\mathrm{this}\:\mathrm{holds}\:\mathrm{for}\:{n}\geqslant\mathrm{2} \\ $$

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