(160+1)^n-1=160^n+C(n,1)*160^(n-1)+...+C(n,i)*160^(n-i)+...+160*n
we can prove C(n,i)*160^(n-i) has more factors of 2 than 160*n. Therefore, 160*n has to be divisible by 2^2005. The smallest n to satisfy this is 2^2000.
C(n,i)*160^(n-i)=C(n,i)*160^i=n*160^i/i*((n-i+1)*...*(n-i)/(1*2*...*(i-1))
The product of (i-1) consecutive integers starting from (n-i+1) has at least as many factors of 2 as the product of (i-1) consecutive integers starting from 1. 160^i has more factors of 2 than i.
we can prove C(n,i)*160^(n-i) has more factors of 2 than 160*n. Therefore, 160*n has to be divisible by 2^2005. The smallest n to satisfy this is 2^2000.
C(n,i)*160^(n-i)=C(n,i)*160^i=n*160^i/i*((n-i+1)*...*(n-i)/(1*2*...*(i-1))
The product of (i-1) consecutive integers starting from (n-i+1) has at least as many factors of 2 as the product of (i-1) consecutive integers starting from 1. 160^i has more factors of 2 than i.
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