# Find a prime factor of $7999973$ without a calculator

How would you go about finding prime factors of a number like $7999973$? I have trivial knowledge about divisor-searching algorithms.

• This is the only context you have to solve the problem. Find a prime factor of 7999973 without a calculator. I got this problem from an old problem book, but there is no explanation or further context.
– user523628
Commented Jan 21, 2018 at 23:52
• Perhaps you didn't read the question carefully... or perhaps you didn't write the question carefully. Finding prime factors (implicitly: all prime factors) of the number, without a calculator, is going to be quite hard. Finding a prime factor (that is: only one asked for), as in your title, is quite easy in this case. Commented Jan 21, 2018 at 23:57
• If ever I could understand what is special with this question and such votes Commented Jan 22, 2018 at 21:27
• Those who don't understand this problem might understand it more if they assumed that it's a problem that is meant to be solvable in a reasonable amount of time. When you take that approach, it's safe to assume there must be some trick or non-linear thinking, as the answer confirms. Commented Jan 22, 2018 at 21:34
• Possible duplicate of How to factorise large number without calculator?
– Rob
Commented Jan 23, 2018 at 10:55

## 1 Answer

The thing to notice here is that 7,999,973 is close to 8,000,000. In fact it is $8000000 - 27$. Both of these are perfect cubes. Differences of cubes always factor: $$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$

Here we have $a=200, b=3$, so $a-b= 197$ is a factor.

• @Bernard: Also note that this method does not prove that the factors we got are prime. Commented Jan 22, 2018 at 9:30
• @user21820 As clarified in a comment, the original question only required finding a prime factor, and 197 is small enough to check by hand.
– G_B
Commented Jan 22, 2018 at 10:26
• @GeoffreyBrent: Yup I know that 197 is prime; I'm just stating for the record that this method in general does not say anything about primality. Commented Jan 22, 2018 at 10:33
• @Rafalon trying to divide by 7, 11 and 13 is enough (for 197). Commented Jan 22, 2018 at 20:02
• @PeterH Because the original number was suspiciously close to $8\,000\,000$, with all those $9$'s, so he checked to see whether it worked, and it did. It's a standard type of problem, although in my experience, powers of $10$ are more common as starting points for these problems than numbers like eight million. Commented Jan 23, 2018 at 9:54