How to compare products of prime factors efficiently? Let's say that $n$ and $m$ are two very large natural numbers, both expressed as product of prime factors, for example:
$n = 3×5×43×367×4931×629281$
$m = 8219×138107×647099$
Now I'd like to know which is smaller. Unfortunately, all I have is an old pocket calculator that can show at most (say) ten digits. So while there are enough digits to enter each factor individually, $n$ and $m$ are both too large to fit in the calculator. To my disappoint, they are also so close that even their logarithms are indistinguishable looking at the first 10 digits.
Question: how would one go to determine which one of two integers is smaller in a case like this? Any easier alternative than calculating the full decimal expansion of both products with pen and paper?
 A: Your pocket calculator can work with the factors of $n$ and $m$, in this case, let $n = n_1\cdot n_2\cdot\ldots n_r$ and $m = m_1\cdot m_2\cdot\ldots m_s$, now calculates $\log(n) = \log(n_1) + \log(n_2) + \ldots + \log(n_r)$ and $\log(m) = \log(m_1) + \log(m_2) + \ldots + \log(m_s)$. This calculations are pretty possible if their factors aren't huge.
Note that $n > m \iff \log(n/ m) > 0$, and $n<m \iff \log(n/ m) < 0$. Therefore all you need to do is check the sign of $\log(n/m)$. 
You have that $\log(n/ m) = \log(n_1) + \ldots + \log(n_r) - \log(m_1) - \ldots - \log(m_s)$, so you can expect to do this test to really large (but not arbitrarily large) numbers.
$\textbf{Edit}$: I just realize that $\log(n/ m) = 
-0.0000000000000027\ldots$ in your example, so maybe this test is not good enough. But the problem are not the zeros (which are more then 10), the problem is if your pocket calculator will return the minus sign or not. Because an output like $-0.0000000000$ would be enough to know which one is bigger.
A: Scientific notation maybe very useful for analysis here.
Notice how for n we have an order of $3 \cdot 5 \cdot 4.3 \times10^1 \cdot 3.7 \times10^2 \cdot 4.9 \times 10^3 \cdot 6.3 \times10^5$
By the properties of exponents and multiplication you know this has a magnitude of at least $10^{11}$ by just adding up the exponents of the tens.
Similarly for m we get a magnitude of at least $10^{13}$ via the same method.
Without using a calculator or multiplying the digits we already get the implication that the m is larger.
However in the interest of precision: multiplying the digits show that both quantities have the exponential magnitude of $10^{14}$, yet the digits of n multiply to roughly 7.4 (after rounding) and the digits of m multiply roughly to 7.5 (after rounding) so the second is probably larger.
Feel free to use as many significant figures for more accuracy.
A: As a practical example, in a programming language like C or C++, you can easily calculate both products using floating-point arithmetic, and you can easily calculate both products modulo 2^64 using unsigned integer arithmetic. 
If you can estimate the rounding error in the floating-point arithmetic products, and prove that the total rounding error is less than 2^63, then it's easy: Calculate the difference d between the products in floating-point arithmetic. If the difference is >= 2^63 or <= -2^63 then that decides. Otherwise, you know that the exact difference is greater than $d - 2^63$ and less than $d + 2^63$. We also know the exact difference modulo 2^64; this is enough to determine the exact difference. This should work for numbers up to 33 or 34 digits. 
