- #1
roam
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Homework Statement
http://img200.imageshack.us/img200/7097/99175506.gif [Broken]
Homework Equations
The Attempt at a Solution
For [tex]n \in N[/tex] let P(n) be the statement: "81 | (10n+1-9n-10)"
Base Case: when n=1: 10n+1-9n-10 = 81 = 81 × 1
So P(1) is true
Inductive Step: let [tex]k \in N[/tex] and suppose [tex]P(k)[/tex] is true, that is 81 | (10k+1-9k-10) is true. Then [tex]10^{k+1}-9k-10=81m[/tex] for some [tex]m \in Z[/tex]. Then [tex]10^{k+1}=(81m+9k+10)[/tex].
So,
10k+2-10(k+1)-10=10 × 10k+1-10(k+1)-10
=10 × (81m+9k+10)-10k+10-10
= 10 × (81m+9k+10-k)
I'm stuck here and I don't know how to factor things out and bring the 81 forth to get the expression in the form: "81(something here)" to show that it's divisible by 81 for all n. Any help is appreciated.
P.S.
I might be able to do it this way:
10 × (81m+9k+10-k) = 810m+90k+100-10k
=> 81(10m+(90/81)k+(100/81)-(10/81)k)
But I don't feel that this is the right way...
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