Ask your own question, for FREE!
Mathematics 51 Online
OpenStudy (anonymous):

smallest positive integer in the set = {6x+15y | x,y integers}

OpenStudy (anonymous):

21xy? I can't really understand what you mean..

OpenStudy (fibonaccichick666):

My guess is that it is asking what the minimum value of that quantity is.

OpenStudy (fibonaccichick666):

can you clarify?

OpenStudy (anonymous):

Oh ok well, idk o.o sorry

OpenStudy (anonymous):

the set is such that it contains all values described by 6x+15y where x and y are integers, for example if x=y=1 then the number would be 6(1)+15(1) = 21, so 21 is in the set, if x= -1 and y= 2 then the number would be 6(-1) + 15(2) = 24 so 24 is in the set, what is the smallest positive integer in the set?

OpenStudy (fibonaccichick666):

smallest positive, that changes things, ok, so, let's see

OpenStudy (fibonaccichick666):

wht have you tried so far mick?

OpenStudy (anonymous):

i've tried "im pretty sure its 3" so far, but i cant justify it

OpenStudy (fibonaccichick666):

so, that sounds reasonable to me, now just so I know what I can use here, what level class is this? High school, college?

OpenStudy (anonymous):

Can 0 be an interger caus it know it means a number, so if it was 6(0) + 15(0) = 21

OpenStudy (anonymous):

sorry i*

OpenStudy (fibonaccichick666):

0 isn't usually considered a positive integer, it's sort of on it's own, but if it was included, that would work, (i think this question implies a>0 )

OpenStudy (fibonaccichick666):

Very nice thought process trin

OpenStudy (anonymous):

its a second year university number theory course

ganeshie8 (ganeshie8):

Hint : \(\gcd(a,b)\) is the smallest positive integer that can be written as \(ax+by\)

OpenStudy (fibonaccichick666):

So now, mick, the first thing I'm gonna hint at is, can you factor something out of 6x +15y?

OpenStudy (anonymous):

hmm. oh, ok hmm. and thanks. Before I go I would look up smallest positive integers and find one that fits! Im only in 6th grade and is advanced but I will try to figure it out :)

OpenStudy (fibonaccichick666):

Keep it up trin! That was an excellent attempt

OpenStudy (anonymous):

But 0 sounds perfect because it is the only smallest positive number and like u said sometimes it can be counted as but not usually... try 1? maybe and Thanks! Gtg for tonight

OpenStudy (anonymous):

yeah gcd(6,15) = 3 like @ganeshie8 said

OpenStudy (fibonaccichick666):

alright, so that gives us 3(2x+5y) right?

OpenStudy (anonymous):

Yea, 3 is a thought!

ganeshie8 (ganeshie8):

+1 for the 6th grader fascinated by numbers xD

OpenStudy (fibonaccichick666):

so now, what is the smallest you can make (2x+5y)

OpenStudy (anonymous):

1

OpenStudy (fibonaccichick666):

right so the smallest 2x+5y can be is 1, now we factored out that 3 remember? so we can write that as 3(2x+5y)--> 3(1) which gives you your smallest possible value :)

OpenStudy (fibonaccichick666):

we can't minimize 3

OpenStudy (fibonaccichick666):

we can only minimize the parentheses, and from the given we can quickly find the minimum positive value there

OpenStudy (fibonaccichick666):

as you did, so plug chug and direct proof accomplished :)

OpenStudy (fibonaccichick666):

They are all prime too which makes this awesome!!!!! Just a side note :D

OpenStudy (anonymous):

ok thanks,

OpenStudy (fibonaccichick666):

np, any questions?

ganeshie8 (ganeshie8):

how do we know the minimm value of 2x+5y is 1 ?

OpenStudy (fibonaccichick666):

x=-2 y=1, minimum integer >0

ganeshie8 (ganeshie8):

right, how did we figure out x = -2 and y=1 ?

OpenStudy (fibonaccichick666):

I considered that obvious, I should have proven that, good catch

OpenStudy (anonymous):

i don't think you have to prove 1 is the smallest positive integer

OpenStudy (fibonaccichick666):

He wants us to show how we got -2 and 1, there's a way to show that

ganeshie8 (ganeshie8):

interesting, so are working it without usign gcd properties.. nice :)

OpenStudy (fibonaccichick666):

(that was my first instinct here... I like simple haha)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!