Find the number of positive integers that divide (10)^999 but not (10)^998.
Let us consider two positive integers A and B with A > B.
Now, we know that the prime factorization of 10^M for any positive integer M is given by:(2^M)*(5^M)
Let is define s(N) # divisors of N that are <= N
Then, we must have:
s(10^A) = s(2^A*5^A) =(A+1)(A+1)=(A+1)^2
Similarly, s(10^B) =(B+1)^2
Accordingly, number of divisors of 10^B that are NOT divisors of 10^A is equal to (A+1)^2-(B+1)^2
Therefore, substituting (A, B)= (999,998), wehave:
# divisors of 10^998 that are NOT divisors of 10^999 is equal to:
1000^2-999^2= (1000+999)(1000-999)= 1999
Puzzle Solution
