How many positive integers are not divisble by 12 or 15?

[1+1=1]

hi i am new to THIS place here but i do put posts on the number theory site as well. i am in need of direction and have no idea where to turn. i need help w/ two ?'s and they are...

how many pos int. <1000 are NOT divisible by 12 or 15?

prove the if the sum of two consec. int. is a perfect sq., then the smaller of the two int. is one side of a prim. right. tri. and the large number is the hypot.

here's my thoughts so far...

as far as the pos. int. go, could i look at all of the cases of 12 and 15? i mean could i say 15k...15k+14 and then say 12k...12k+11 or would that be too long and drawnout? anyone with a simpler idea?

as far as the perfect sq, i know that by consec. int. means i can say this... n is one int. and n+m is the second consec. int. so, i'd have n,n+m. so if the larger ot the two is the hypot. then i can say n + ____ = n+m, right? anyone help with these two would be greatly appreciated! also, let me know if i am thinking correctly! thanks you all!

 

[Muzza]

Seriously man, your posts would be much more readable if you didn't use all those weird abbreviations.

how many pos int. <1000 are NOT divisible by 12 or 15?

Consider the easier problem of counting the number of integers < 1000 that ARE divisible by 12 or 15. Those numbers + the number of integers NOT divisible by 12 or 15 = 1000...

prove the if the sum of two consec. int. is a perfect sq., then the smaller of the two int. is one side of a prim. right. tri. and the large number is the hypot.

Right, first you have to prove that they form the sides of a right triangle with integer lentgths. You know that (n) + (n + 1) = m^2 for some m. Is there an integer x such that n^2 + x^2 = (n + 1)^2 (Pythagorean theorem)? If you solve for x^2, you'll probably be pleasantly surprised... Then you've got to prove that their gcd is 1, but take one step at a time ;)

 

[1+1=1]

y does it seems simpler when you explain things to me?

 

[1+1=1]

muzza, how would i check the gcd = 1? i got the formula to work out, i just need that and that is it.

 

[Muzza]

I'm not sure actually, I thought it'd be straight forward, but it isn't :P It's that bloody square root that messes things up...

 

[1+1=1]

you';re tellin me. haha i just got it to the form that is discussed. i don't know if my teacher would care about gcd and all that.

 

[1+1=1]

i'm still stuck on the < 1000 thing. i don't know what you mean when you say " Consider the easier problem of counting the number of integers < 1000 that ARE divisible by 12 or 15. Those numbers + the number of integers NOT divisible by 12 or 15 = 1000..." i just am lost as to how you can come up with that? can anyone explain??

 

[AKG]

how many pos int. <1000 are NOT divisible by 12 or 15?

As Muzza said, it would be nice if you didn't use abbreviations. In fact, you're here to ask for help, and often it's tough having to read your posts, or requires two reads. Not a big deal, but considering the fact that you're asking for help, I don't think it would hurt to take the time to make it nice and easy to read. Grammar, spelling, organization (i.e. numbering questions maybe) would be great.

There are 999 positive integers less than 1000. How many are not divisible by 12 or 15? Well, this is equal to 999 - "the number of positive integers less than 1000 divisible by 15"(a) - "the number of positive integers less than 1000 divisible by 12"(b) + "the number of positive integers less than 1000 that divisible by 12 and 15"(c). To be divisible by 12 and 15, it must be divisible by their least common multiple, 60. It's easy now:

a = \left\lfloor{999/15}\right\rfloor = \left\lfloor{66.6}\right\rfloor = 66

b = \left\lfloor{999/12}\right\rfloor = \left\lfloor{83.25}\right\rfloor = 83

c = \left\lfloor{999/60}\right\rfloor = \left\lfloor{16.65}\right\rfloor = 16

So, the answer you want is 999 - 66 - 83 + 12 = 862. There are 862 positive integers less than 1000 that are not divisible by 12 or 15. And if you're not totally convinced that the number you're looking for is 999 - a - b + c, draw a Venn diagram.