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Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion


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December 18th, 2016, 07:43 AM   #11
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Thanks again for your excellent solution, romsek.
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January 9th, 2017, 11:15 PM   #12
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Here's the number theory approach!

You must know that there are 25 primes less than 100 and they are:
2 | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 29 | 31 | 37 | 41 | 43 | 47 | 53 | 59 | 61 | 67 | 71 | 73 | 79 | 83 | 89 | 97 (25 primes)

For a number N to have exactly four factors (divisors), the number should be of the form N = p³ or pq where p and q are distinct prime numbers.

Case 1: N =p³ form
Clearly, p can acquire only 2 values: 2 and 3

So, there are a total of 2 numbers of the form N =p³.

Case 2: N =pq form such that q>p

When p=2, 0<q<50
So, there are 14 values that q can possibly have.

When p=3, 0<q<33.33
So, there could be possibly 9 values for q.

When p=5, 0<q<20
So, there are only 5 values of q.

When p=7, 0<q<14.28
So, there are 2 values of q.

We should not move ahead!

Therefore, there are a total of 14 + 9 + 5 + 2 = 30 numbers of the form N =pq

Thus, there are 2 + 30 = 32 numbers that have exactly 4 factors (divisors).

I hope it helps!

Last edited by skipjack; January 10th, 2017 at 11:02 AM.
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