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Question Number 200041 by depressiveshrek last updated on 12/Nov/23
Bystronginductionprovethatanynaturalnumberequaltoorbiggerthan8canbewrittenas3a+5bwhereaandbarenon−negativeintegers.
Answered by des_ last updated on 12/Nov/23
n⩾8⇒n=3a+5b,a,b⩾0;1)n=8:n=3(1)+5(1)2)n⇒n+1:2.1)n=3a+5b,b>0:n+1=3a+5b+1=3(a+2)+5(b−1);2.2)n=3a+5b,b=0:n=3a⇒a⩾3;n+1=3a+1=3(a−3)+5(2);Thusn⩾8⇒n=3a+5b,a,b⩾0
Answered by witcher3 last updated on 12/Nov/23
showforn∈[8,16]..Bytchekingforall16⩽k⩽n⇒P(k)Truep(n+1)n+1=[n+12]m+(n−[n+12])=d8⩽m,d<nm=3a+5dd=3a′+5d′m+d=n=3(a+a′)+5(d+d′)
wecanshowthiseaslyn=8k+rk⩾1forn⩾8r=0,3.0+5.0r=1=3.2+5(−1)r=25.1+3(−1)r=3=3.0r=4=3.3+5(−1)r=5=5.1+3.0r=6,3.2+5(0)r=7,5.2+3(−1)∀r∈[0,7]r=3.a+b.5min(a,b)⩾−1n=k(3+5)+3a+5b=3(a+k)+5(k+b)a+k⩾k−1⩾0b+k⩾k−1⩾0..True
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