Suppose that cats are only sold in batches of 3 or k. Is it ways true that there’s some n0 such that if n ≥ n0, you can purchase exactly n cats? When such an n0 exists, how does it depend on k? What if cats were sold in batches of a or b?

It isnt always true. It is only true when k is not a multiple of 3. If cats were sold in batches of a and b, then they have to be coprime, in other words, its only positive common divisor should be 1.

Step-by-step explanation:

If k is a multiple of 3, then any combination of batches you bought will give as a result a multiple of 3. Thus, you cant but, lets say 31 cats, or 301, or 3001, and so on.

If k is not a multiple of 3. Then k and 3 are coprime, which means that there exists n and m such that 3n + mk = 1.

Thus,

3n + mk = 1

6n + 2mk = 2

One either m or n is negative. If, for example, n is negative, then, we will be able to form any number from -3*2n = -6n (which is positive) onwards, because

-6n + 1 = -3*2n + (3n+mk) = 3*(-n) + mk

-6n + 2 = -6n + (6n + mk) = mk

And any other number greater than -6n+2 is obtained either from 6n, 6n+1 or 6n+2 by adding a positive multiple of 3.

For m negative the argument is similar.

If cats were solver in batches of a or b, then we can only get cats that are a multiple of the greater common divisor of a and b. If that greater common divisor is 1 (in other words, a and b are coprime), then, we can obtain any number large enough.

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.

the answer is mp

step-by-step explanation: