Prime factorization of a factorial

Question

I need to write a program to input a number and output its factorial\'s prime factorization in the form:

4!=(2^3)*(3^1)

5!=(2^3)*(3^1)*(5^1)

Answer 1

2^3 is another way of writing 2³, or two to the third power. (2^3)(3^1)(5^1) = 2³ × 3 × 5 = 120.

(2^3)(3^1)(5^1) is just the prime factorization of 120 expressed in plain ASCII text rather than with pretty mathematical formatting. Your assignment requires output in this form simply because it's easier for you to output than it would be for you to figure out how to output formatted equations (and probably because it's easier to process for grading).

The conventions used here for expressing equations in plain text are standard enough that you can directly type this text into google.com or wolframalpha.com and it will calculate the result as 120 for you: (2^3)(3^1)(5^1) on wolframalpha.com / (2^3)(3^1)(5^1) on google.com

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WolframAlpha can also compute prime factorizations, which you can use to get test results to compare your program with. For example: prime factorization of 1000!

A naïve solution that actually calculates the factorial will only handle numbers up to 12 (if using 32 bit ints). This is because 13! is ~6.2 billion, larger than the largest number that can be represented in a 32 bit int.

However it's possible to handle much larger inputs if you avoid calculating the factorial first. I'm not going to tell you exactly how to do that because either figuring it out is part of your assignment or you can ask your prof/TAs. But below are some hints.

a^b × a^c = a^b+c

equation (a)      10 = 2¹ × 5¹
equation (b)      15 = 3¹ × 5¹
10 × 15 = ?      Answer using the right hand sides of equations (a) and (b), not with the number 150.

10 × 15 = (2¹ × 5¹) × (3¹ × 5¹) = 2¹ × 3¹ × (5¹ × 5¹) = 2¹ × 3¹ × 5²

As you can see, computing the prime factorization of 10 × 15 can be done without multiplying 10 by 15; You can instead compute the prime factorization of the individual terms and then combine those factorizations.