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f(N)=[N/1] + [N/2] + [N/3]+ ....+ [N/N] . Find f(64)-f(63).
If you could reason carefully you see f(N+1)-f(N)=no. of factors og N+1. Because only for these numbers(factors of N+1) ; [N+1/these no.]=[N/these no.]+1. You could verify it for any no. N. f(64)-f(63)=7 ;{factors of 64 are 1,2,4,8,16,32,64}.
If you could reason carefully you see f(N+1)-f(N)=no. of factors og N+1.
Because only for these numbers(factors of N+1) ; [N+1/these no.]=[N/these no.]+1.
You could verify it for any no. N.
f(64)-f(63)=7 ;{factors of 64 are 1,2,4,8,16,32,64}.
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