Find the number of ordered pairs $(m, n)$ of positive integers such that $mn = 2010020020010002$

Question

No calculators allowed, so I don't know how to approach this problem, and I wonder if there's an easier way than factoring it all out by hand.

Answer 1

As Michael T noticed, you need to find the number of divisors of your number and to do so, you can fin the prime factorization of the number. Looking at the number it is clearly divisible by $2$ so $$k = 2010020020010002 = 2 \times1005010010005001$$ Now, the last factor present a interesting patern. if you look at the numbers that apear at the position multiples of 3 (starting with 0) you have $$1 \ \ 5 \ \ 10 \ \ 10 \ \ 5 \ \ 1$$ tha are combinatorial numbers(a row of Pascal triangle) so $$1005010010005001 = \binom{5}{5} (1000)^5 + \binom{5}{4} (1000)^4 + \binom{5}{3} (1000)^3 + \binom{5}{2} (1000)^2 + \binom{5}{1} (1000)^1 + \binom{5}{0} = (1000 + 1)^5$$ by Binomial theorem and $1001 = 7\times 11\times 13$. So $$k = 2\times(1001)^5 = 2\times 7^5 \times 11^5 \times 13^5$$

Answer 2

$$2010020020010002=2^{1}*7^{5}*11^{5}*13^{5}$$,then $answer=(1+1)*(5+1)*(5+1)*(5+1)=432$