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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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January 21st, 2017, 06:12 PM   #13
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The problem is:

Quote:
Is there a way to find the positive integers with 4 factors less than 100?
I understand this to mean:

Quote:
Is there a way to find the positive integers with 4 factors each of which is less than 100?
So why are we adding the factors up? Do you understand it to mean the the sum of the factors must be less than 100?
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