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Question Number 207864 by efronzo1 last updated on 29/May/24

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Answered by Rasheed.Sindhi last updated on 29/May/24

2^(5m) ∙5^(2n) ∙k=2020^(2020) =(2^2 .5.101)^(2020) 2^(5m) =2^(4040) ∧ 5^(2n) =5^(2020) ∧k=101^(2020) 5m≤4040 ∧ 2n≤2020 m_(max) =((4040)/5)=808 ∧ n_(max) =((2020)/2)=1010 max(n+2m)=n_(max) +2m_(max) =1010+2(808)=2626

25m⋅52n⋅k=20202020=(22.5.101)202025m=24040∧52n=52020∧k=10120205m⩽4040∧2n⩽2020mmax=40405=808∧nmax=20202=1010max(n+2m)=nmax+2mmax=1010+2(808)=2626

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