Problem with sum of factors.

Question

The sum of the total number of factors of $999000$, $816480$ and $819529$ is $n$. How many ways can $n$ be written as $\sqrt{a}+b$ where $b$ is a non-negative positive integer?

Answer 1

The prime factorization of all of your numbers is: $$999000= 2^\color{red}{3}\cdot3^\color{red}{3}\cdot5^\color{red}{3}\cdot37^\color{red}{1},$$ $$816480=2^\color{red}{5}\cdot3^\color{red}{6}\cdot5^\color{red}{1}\cdot7^\color{red}{1},$$ $$819529=743^\color{red}{1}\cdot1103^\color{red}{1}.$$

Now we can use the formula for the number of factors of a number, which is to take all of the exponents of the prime factors, add $1$ to each, and multiply them together.$^1$ So the number of factors of each number is $$999000:(\color{red}{3}+1)(\color{red}{3}+1)(\color{red}{3}+1)(\color{red}{1}+1)=128,$$ $$816480:(\color{red}{5}+1)(\color{red}{6}+1)(\color{red}{1}+1)(\color{red}{1}+1)=168,$$ $$819529:(\color{red}{1}+1)(\color{red}{1}+1)=4.$$

Therefore, the sum of these numbers is $n=128+168+4=300.$ Now, we need to find the number of ways to express this in the form $\sqrt{a}+b.$ Since $b$ is a positive integer, if $a$ is not a perfect square, then $\sqrt{a}$ will be irrational, and then $\sqrt{a} + b$ will be irrational, so $\sqrt{a} + b$ will be only be $300$ if $a$ is a perfect square. Now we just need to count the number of perfect squares $a$ such that $\sqrt{a} < 300$. This is all of the perfect squares from $0^2$ to $299^2$, which is $\color{red}{300}$.

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$^1$If you're interested in where this formula comes from, it's just from combinatorics. Look at the number $2^3$, for example. The number of factors of this are $2^0,\ 2^1,\ 2^2,$ and $2^3$. And in general for any number $p^n$ where $p$ is prime there will be $n+1$ factors of this number: $p^0,$ $p^1,$ $p^2,$ ..., $p^n.$ Now, if we have a product of prime numbers $n = p_1^{a_1}p_2^{a_2} \cdots p_n^{a_n}$, the $n$th prime has $a_n + 1$ factors as we showed before, and any combination of factors from any of these primes will be a factor of $n$. So therefore the total number of factors you can produce is given by the product $(a_1 + 1)(a_2 + 1)\cdots(a_n + 1)$.

Answer 2

Hint: you are being blinded by large numbers and confusing symbols to a really simple problem. The only two issues to show that $c=\sqrt a$ is an integer, and whether $a$ is allowed to be zero.

Express the problem in terms of $c$ ...