Equation integer part
what are all the steps to solve this equation?
97645643 – 6X * lower part( 97645643 / 6X)=X
thank you
97645643 – 6X * lower part( 97645643 / 6X)=X
thank you
If x is assumed to be an integer, you can move the 6x inside of the floor operation. This then cancels the 6x in the denominator. You are left with 97645643 - 97645643 = x
so x = 0
I'm not sure if this is a legitimate solution as this implies that 0 is in the denominator.
so x = 0
I'm not sure if this is a legitimate solution as this implies that 0 is in the denominator.
Original post by morgan8002
If x is assumed to be an integer, you can move the 6x inside of the floor operation. This then cancels the 6x in the denominator. You are left with 97645643 - 97645643 = x
so x = 0
I'm not sure if this is a legitimate solution as this implies that 0 is in the denominator.
so x = 0
I'm not sure if this is a legitimate solution as this implies that 0 is in the denominator.
This won't yield integer solutions for x.
Rearranging..
Unparseable latex formula:
\dfrac{97645643}{x}\right \rfloor =1+6 \left\lfloor \dfrac{97645643}{6x}\right \rfloor
So x is a factor of 97645643.
Original post by gerva
but what are all the values of x, and how to calculate
As argued above, x is an integer which is 5 mod 6, which divides that long number. Sadly it's hard to factor that number.
Notice that is "a rounded down to the nearest multiple of b". Hence is "the difference between a and the nearest multiple of b less than a": that is, it is .
That is, we seek such that .
Does that help? (You may find the Chinese Remainder Theorem useful in seeing from another angle where this came from.)
(edited 9 years ago)
Original post by gerva
yes, thank you
Does it? I can't really see that it does, but if it helped… :P Mind writing out your solution, because I still haven't solved it except by computer?
Original post by Smaug123
Does it? I can't really see that it does, but if it helped… :P Mind writing out your solution, because I still haven't solved it except by computer?
This one, by luck, comes out very quickly using Fermat's factorisation method.
if you want to see what I'm doing http://albericolepore.altervista.org/
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