find primes P1, P2, P3 such that P1*P2*P3 = 190747

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twisted_logic89

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P1*P2*P3 = 190747

what are the numerical values for P1, P2, P3 (they are prime numbers)

um... someone actually took the time to do the trial and error way, and found the numbers to be 53, 59, 61
what is the mathematical way of finding something like this out?
 
Re: P1*P2*P3 = 190747

Not sure if this is what you're looking for. I'm unaware of any theorems/formulae that will spit these out. If they are assumed distinct primes, then at least one of them is larger than or equal to 7. Say our product of primes is p=x*y*z, and we are assuming that x >= 7. (2 clearly does not work)

Therefore p = x*y*z >= 7*y*z which implies y*z <= p/7. But we only need to consider the integer part, [p/7].

In our case, we have y*z <= [190747/7] = 27249. A bit smaller... fixed mistake: changed 5 to 7

Now, we can simplify this a bit further, since we need only to check until the number [sqrt(27249)] = 165 to find ONE of the primes.

Knowing that we only need to check odds, this comes out to about 82 iterations. The higher prime you check, the smaller this number becomes, so a dynamic algorithm here would be much faster. i.e. if you check that no primes up to 11 divide it, you can assume at least one prime is larger than or equal to 17, and divide by 17 above instead of 7 (about 50 iterations).

If one of the primes is small, with a large number this becomes less useful, however.
 
Re: P1*P2*P3 = 190747

I appreciate your help, but some of the things you posted have me a little confused...

In our case, we have y*z <= [190747/5] = 27249
Click to expand...

190747/5 is 38149.4..........

Now, we can simplify this a bit further, since we need only to check until the number [sqrt(27249)] = 165 to find ONE of the primes.
Click to expand...

sqrt of 27249 is 165.0727113, not 165 even.....
and I just realized, did you mean to say 190747/7 instead of 190747/5? If so, 190747/7 is 27249.57143, not 27249 even...

???
 
Re: P1*P2*P3 = 190747

twisted_logic89 said:
um... someone actually took the time to do [the math] ... what is the mathematical way of finding something like this out?
Click to expand...


There are less than twelve divisions needed; perhaps, these divisions comprise the mathematical way of finding out something like "this".

 
Re: P1*P2*P3 = 190747

twisted_logic89 said:
... what do I divide and by what?
Click to expand...


You could divide 190747 by prime numbers, beginning with 17.

 
Re: P1*P2*P3 = 190747

sqrt190747 = 436.75...
190747/ 17 = 11220.41...
/19= 10039
/23
/29
/31
/37
41
43
47
53
59
61
67
71
73
79
83
89
97
101
103
107
109
113
127
131
137
139
149
151
157
163
167
173
179
181
191
193
197
199
211
223
227
229
233
239
241
251
257
263
269
271
277
281
283
293
307
311
313
317
331
337
347
349
353
359
367
373
379
383
389
397
401
409
419
421
431
433

WOW...... in that list of prime numbers up to the sqrt of 190747, I'm noticing most dont come out even. Except for 53, 59, and 61. So is that why they are the numbers, because they are the only primes that evenly divide it?
 
Re: P1*P2*P3 = 190747

twisted_logic89 said:
P1*P2*P3 = 190747

what are the numerical values for P1, P2, P3 (they are prime numbers)

um... someone actually took the time to do the trial and error way, and found the numbers to be 53, 59, 61
what is the mathematical way of finding something like this out?
Click to expand...

In the problem - did they state that these are consecutive primes?
 
Re: P1*P2*P3 = 190747

twisted_logic89 said:
I appreciate your help, but some of the things you posted have me a little confused...

In our case, we have y*z <= [190747/5] = 27249
Click to expand...

190747/5 is 38149.4..........

[quote:3q95ff8z]Now, we can simplify this a bit further, since we need only to check until the number [sqrt(27249)] = 165 to find ONE of the primes.
Click to expand...

sqrt of 27249 is 165.0727113, not 165 even.....
and I just realized, did you mean to say 190747/7 instead of 190747/5? If so, 190747/7 is 27249.57143, not 27249 even...

???[/quote:3q95ff8z]


yes, I meant 7.

The non-integer part is useless. The [ and ] are notation for the "greatest integer function". if y*z <= 27249.57 then both y and z must divide an integer <= 27249. (they are prime numbers)

get it?
 
P1*P2*P3 = 190747
what are the numerical values for P1, P2, P3 (they are prime numbers)
um... someone actually took the time to do the trial and error way, and found the numbers to be 53, 59, 61
what is the mathematical way of finding something like this out?

Something I have used for years with reasonably high success.

Since we are dealing with 3 numbers, take the cube root of 190747 = 57.564

Being greater than 57, I move to the next prime number 59 as the middle number.

190747/59 = 3233.

Seeking 2 other numbers on both sides of 59, take the square root of 3233 which is 56.859.

The closest prime below 56.859 is 53.

The closest prime above 56.859 (excluding 59) is 61.

Thus, 53x59x61 = 190,747.

The method works best when the three numbers are relatively close to one another. i.e., consecutive or having a small constant difference. (Thanks to Subhotosh Kahn)
 
The procedure above will work well - if the numbers are "reasonably" equally spaced.

If suppose the number was 11*59*61 - the procedure above becomes cumbersome.
 
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