How many functions such that:
for all and ,
for any whose last digit is 3, .
Classification:
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Since f 6 ( 2 ) f 2 ( 3 ) f ( 5 ) f ( 7 ) = f ( 2 0 1 6 0 ) = 1 we must have f ( 2 ) = f ( 3 ) = f ( 5 ) = f ( 7 ) = 1 .
Since f ( 2 ) = f ( 2 × 1 ) = f ( 2 ) f ( 1 ) we also have f ( 1 ) = 1 .
If n is a positive integer that is not a multiple of 2 or 5 then for some m , n m has last digit 3.
Therefore f ( n ) f ( m ) = f ( n m ) = 1 and hence f ( n ) = 1 .
If n = 2 a 5 b n 0 where n 0 is not a multiple of 2 or 5 then
f ( n ) = f a ( 2 ) f b ( 5 ) f ( n 0 ) = 1
So f ( n ) must equal 1 for all non-zero n .
Finally note that f ( 0 ) = f ( 0 ) f ( 0 ) so f ( 0 ) = 0 or f ( 0 ) = 1 .
So, there are 2 functions such that all of above conditions.