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/sci/ - Science & Math

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4678107 No.4678107 [Reply] [Original]

how do i make wolfram know i want an integer solution?

i tried
(2+3k)/4 = floor((2+3k)/4), k is integer

he doesnt seem to understand

>> No.4678116

if k = 4 (2+3k)/4 = 7/2

floor(7/2) != 7/2

the relation is false

>> No.4678117
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4678117

wat?
if the solution is an integer, it will give it
if its not, its just not!
wolfram aint gonna lie, nigga

>> No.4678143

>>4678107
You could round it to become an integer, don't know how
to do that though...

>> No.4678144

Ummm, if you only want to return the integer part of a solution just use floor(7/2) as an example.

Shames me to say it but I agree with EK

>> No.4678158

(2+3k)/4 = floor((2+3k)/4), solve for k

That should work

>> No.4678170

all interger solutions are in the form of
4n-2=k for every integer n>=1
for example n=2 so k=6 so both functions equal 5

EK confirmed for retard

>> No.4678179

>>4678170
i didnt actually look at the problem, i was just pointing out that whether the solution is an integer or not depends on what you're trying to solve, it either will be, or it wont. you cant get an integer solution if it just isnt, and vice versa.

>> No.4678180

>>4678179

Fuck you.

>> No.4678183

>>4678117
there are infinite solutions and an infinite set of integer solutions

>>4678144
i dont want the integer part of a solution, i want the solutions which are integer

>>4678158
MAGIC

>> No.4678187
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4678187

>>4678180
woah, whats your problem??

>> No.4678189

>>4678170
actually this works for every n interger positive or negative

>> No.4678198

>>4678170
this works except the range of intergers can also be negative so all intergers between -infinity<n<infintiy are solutions including when n=0

>> No.4678202

im trying to find a method of finding a number that belongs to two arithmetic sequences

like,
c = 2 mod 3 = mod 4
one solution is c=8, the other solutions are c=8+12n

>> No.4678206

>>4678202
this guy solved it
>>4678198
>>4678170

>> No.4678208

>>4678202
how do i congruent signal

>> No.4678217 [DELETED] 

<span class="math">c \cong 2 mod 3 \cong mod 4[/spoiler]

>> No.4678221

<span class="math"> c \equiv 2 mod 3 \equiv mod 4[/spoiler]

>>4678206
for this particular case yes, but not for a general case

>> No.4678241

>>4678221

for every interger n this works >>4678170
I call troll on your stupid ass. Your solution c=8+12n doesn't work since you pin point for what values of n it works for.

2/10 cause I replies

>> No.4678247

>>4678241

you can't* not can

these are right>>4678198

>>4678170

>> No.4678248

>>4678241
thats not what i mean at all
do you guys even read
it feel like i have to make an enormous effort to make myself clear cuz you faggots will misunderstand everything

>> No.4678254

>>4678248
what's your issue with >>4678170

it finds every integer k that is a solution. And yes I know that n can also be negative intergers as well as k. Find a case in which it doesn't.

>> No.4678269

>>4678248
/vent

using the notation i used earlier
the general case would be a,b,c given constants
(a+bk)/c = floor((a+bk)/c), find an integer k for which this is true

and c=8+12n works for every integer n
but it is not a solution for k, it is a solution for 2+3k
the solution k=4n-2 works, 2+3(4n-2) = 2+12n-6 = -4+12n
n=n+1, -4+12n = 8+12n