Math Problem: Factors

39614081257132168796771975168 = 2^95 (95 prime factors, 1 distinct)

It has 96 DIVISORS (factors) and 76 of them are over 1,000,000.  You obtain them as follows:

2^0, 2^1, 2^2, 2^3, 2^4...........and so on until you get to: 2^20 = 1,048,576. So, the powers of 2 from 0 to 19 =20 factors(divisors) that are under 1,000,000.

I'm sorry, I wasn't allowed to use a calculator for this type of problem.

I knew that at some point 2^n was going to be over 1,000,000.

Is there a method to get there without using a calculator?

The problem can be super easy after you find out 2^n is larger than 1,000,000

I suppose you could do it in your head by knowing that 2^10 is just over 1,000. So, 2^20 would be 1,000 x 1,000 +.

The only factors it has are 2s. You can multiply 2 by itself ANY number of times up to 95 times and the results will all be factors of 95.

Yes, that makes sense.

Because 2^10 is slightly over 1000

So 2^20 is kinda slightly over 1,000,000

We know that 2^19 is NOT over 1,000,000 because 2^20 divided by 2 will be significantly under 1,000,000

Factors of 2^95 = {1,2,4,8,16,...,2^94,2^95}.

The first power of 2 that is greater than 1,000,000 is 2^20, which is 1.048,576.

Number of factors of 2^95 which are greater than 1,000,000 = 95 - 20 + 1 = 76