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Today, we are looking at counting a primes exponent inequality problem. Despite all those scary terms, this one is actually fairly straightforward once you master the ability to count prime factors.

Counting primes is all about understanding how many versions of each prime are necessary to construct the entire prime factorization of an integer. In this problem, we are comparing 25s and 5s and we are being asked how many 25s versus how many 5s there are.

Notice how we are not diving into the math immediately. We are first putting this in terms of counting only. 5 to the 12th means that we are actually multiplying 5 by itself 12 times. Like this: 5x5x5x5x5x5x5x5x5x5x5x5. We can now say we have 12 fives. The question then becomes: how many 25s is this equivalent to?

We are now looking for inequality by forming a baseline of equivalents. We now understand how much too many or too few would be. The key question here is how many 5s make up 25? The answer is not 5: we are not dividing or multiplying. 2 prime factors of 5 make 25. 5×5. That is 25=5 square. We wouldn’t know how many 25 it takes to hate more than 12 5s. Where each 25 is the equivalent of 2 5s, 6 25s is the same as 12 5s. So, we need now a 7th 25 in order to have more 5s than the 12 5s on the other side.