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Any math people out there ?
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| `pr0digy |
Okay, so there's one big number, and smaller numbers were multiplied together to get that large number. I know the large number, but not the small numbers. Is there any way to find out the smaller numbers, or do I have to try a brute force type of approach ? Any way to find out anything about the smaller numbers ?
Hopefully I've been concise enough for you to understand what I'm trying to find out. |
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| StereoPrincess |
what's the big number? and how many smaller numbers?
i don't think you have been explanitory enough in your question. :) |
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| cbxzcm |
| quote: | Originally posted by `pr0digy
Okay, so there's one big number, and smaller numbers were multiplied together to get that large number. I know the large number, but not the small numbers. Is there any way to find out the smaller numbers, or do I have to try a brute force type of approach ? Any way to find out anything about the smaller numbers ?
Hopefully I've been concise enough for you to understand what I'm trying to find out. |
Is there some kind of restriction to the small numbers that the problem is looking for? There's an endless amount of combinations of smaller numbers that when multiplied will give you the big number. However, I can tell you that the smaller numbers can't be zero. |
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| Noisician |
in some cases u can use division properties to figure out whether or not your initial number is a multiple of a certain smaller number. for example,
1. a natural number is divisible by 2 if and only if its last digit is even
2. a natural number is divisible by 5 if and only if its last digit is 5 or 0
3. a natural number is divisible by 10 if and only if its last digit is 0
4. a natural number is divisible by 4 (25) if and only if two of its last digits are both 0 or they form a number divisible by 4 (25)
5. a natural number is divisible by 8 (125) if and only if three of its last digits are all 0 or they form a number divisible by 8 (125)
6. a natural number is divisible by 3 (9) if and only if the sum of its digits is divisible by 3 (9)
7. a natural number is divisible by 7 (11 or 13) if and only if the difference between the number formed by that natural number's last three digits and the number formed by the rest of the digits (or vice versa) is divisible by 7 (11 or 13) [for instance, 217184 is a multiple of 11 because 217-184=33 is divisible by 11]
etc. etc. |
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| `pr0digy |
| quote: | Originally posted by StereoPrincess
what's the big number? and how many smaller numbers?
i don't think you have been explanitory enough in your question. :) |
The way I'm trying to figure this out, the bigger number could be anything, and we don't know the amount of smaller numbers.
| quote: | | in some cases u can use division properties to figure out whether or not your initial number is a multiple of a certain smaller number. for example, |
Will these properties still work in this case ? There's multiple numbers being multiplied to get to the big number... |
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| icyhandofcrap |
remember the number is prime if no positive number less than the square root of the number is a factor.
are you trying to find the number of factors? |
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| `pr0digy |
| quote: | Originally posted by icyhandofcrap
remember the number is prime if no positive number less than the square root of the number is a factor.
are you trying to find the number of factors? |
No, not the number of factors... I'll try to explain it better.
Let's say the "big number" is 123456. Now, I know that other numbers were multiplied together to get that big number. So, are there any hints to use in finding out what the smaller numbers are ?
In another, simplified, if the "big number" is 100, there could be 2 "small numbers" of 10 and 10.
Right now I'm writing a program that just multiplies n random numbers together over and over until it gets a winning combo. If there's any "hints" so to speak though, it'd greatly imporve efficiency. (A hint being something like "If [last digit is even] then [multiply only even numbers]").
^Hopefully some of that makes sense. I'm thinking I'll just have to brute force the thing. |
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| Noisician |
| u can transform any integer n>1 to its standard factored form. use the properties i already mentioned in my previous post to find all the factors of your number. u can write a program that keeps factoring the initial number until all its factors are prime numbers. the unique factorization theorem guarantees that it will succeed in a finite number of steps. |
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| `pr0digy |
| quote: | Originally posted by Noisician
u can transform any integer n>1 to its standard factored form. use the properties i already mentioned in my previous post to find all the factors of your number. u can write a program that keeps factoring the initial number until all its factors are prime numbers. the unique factorization theorem guarantees that it will succeed in a finite number of steps. |
Thanks Noisician, this seems to be the exact thing I was looking for :) |
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| icyhandofcrap |
also all you have to do is use a for loop once you get a complete factorization.
like if you had a number that was 2^3 * 3^4
all u have to do is start with 2^0 * 3^(0 .. 4) then to 2^1.. until 3 and you have have all the factors. |
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| getfoul |
in the case of 100
you can do 100= 10 * 10 = 2 * 5 * 2 * 5
to get the multiple numbers, just break it down. |
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| mndeg |
| where is martinique and is everyone there good at math |
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