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How many numbers between 1 and 10,000 have exactly two(2) prime factors, no more and no less? Thanks for any help?

 Apr 15, 2019
 #1
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Not sure about this, but I believe it will be the squares of all primes from 2 to 100

 

So

 

4, 9,25, 49, 121 , 169, 289, 361, 529, 841,961,1369, 1681, 1849, 2209, 2809 , 3481, 3721, 4489, 5041, 

 

5329, 6241, 6889, 7921, 9409

 

I count 25....but....as I did it hurriedly, I might have missed one or two.....better double-check 

 

cool cool cool 

 Apr 15, 2019
 #2
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There are actually quite a few more. This is the procedure used to count them:
1- First, you have to count the number of primes between 2 and 10,000/2, or 5,000, of which there are 669 such prime numbers, since the largest of them,4,999 can be multiplied by the smallest of them, or 2, which will give,9,998 which is the largest such number under 10,000. Then 2 will be multiplied by every prime from 3 up to 4,999.


2- In addition to squaring all 25 primes under 100, then as indicated in (1) above, will continue thus: 2x3, 2x5, 2x7.......etc. till 2 x 4,987, 2 x 4,993, 2 x 4,999. Once we are done with 2, then will continue with 3: 3 x 5, 3 x 7, 3 x 11.....etc. till 3 x 3,323, 3 x 3,329, 3 x 3,331. Then will continue with 5: 5 x 7, 5 x 11, 5 x 13.....etc. till: 5 x 1,993, 5 x 1,997, 5 x 1999 and so on....


3-The above process will continue until we reach the two primes under sqrt(10,000), or thereabout. Actually 97 x 103 =9,991.


4 - The above process can be done accurately by a computer program written specifically for this purpose. Since I have a good knowledge of 2 programming languages, I wrote one just for this specific task.


5 - The final result looks like this:
(4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87......

........ 5006, 5007, 5017, 5027, 5029, 5033, 5041, 5042, 5045, 5053, 5057, 5062, 5063, 5065, 5069, 5071...........

........ 9974, 9977, 9979, 9983, 9985, 9986, 9987, 9989, 9991, 9993, 9995, 9997, 9998.


6 - So, the total number =2,625. Every one of these numbers has ONLY two prime factors by delibrate calculation!.


7 - And that is the END.


 

 Apr 16, 2019

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