Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Then LT(2^5+3^5)=75. Define
to be the smallest integer>5 such that 1) LT(2^
+3^
)=00 or 25 or 50 or 75; 2) LT(2^
)=LT(2^5); 3) LT(3^
)=LT(3^5). Such a
is found by inspecting the first 20 terms of the table to be
. And LT(2^
+3^
)=75. No problem with this?
Now because LT function takes only the last two digits, it is true that LT(2^(5+i))=LT
=LT(LT(2^5)LT(2^i))=LT(LT(2^
)LT(2^i))=LT(
)=LT(
), for any i>0.
Similarly LT(3^(5+i))=LT( ), too. So LT(2^(5+i)+3^(5+i))=LT(LT(2^(5+i))+LT(3^(5+i)))=LT (LT(2^( ))+LT(3^( )))=LT(2^( +i)+3^( +i))
From this we can see f(k)=LT(2^k+3^k) takes only FINITELY MANY values for k=5,6,7,...
and the values it can take are exactly equal to this set:{f(5),f(6),f(7),...,f(
)}, and any f(k)
{00,25,50,75} iff there exists 5
k'
such that f(k')=f(k).
So what is this k' if f(k')
{00,25,50,75}? Inspect the table we find k'=5 or k'=15.
So now you see where k=5+20m and k=15+20m come from.
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If this doesn't convince you. I happily quit
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EDIT: red letters.