Aradio station gives away $15 to every 15th caller, $25 to every 25th caller, and free concert tickets to every 100th caller. when will the station first give away all three prizes to one caller? . show your work

We want to find the lowest common multiple of 3 numbers.

We will see that the 300th caller will get the 3 prizes.

Ok, we know that:

We want to see the first time that the 3 prizes will be given at the same time (at the same caller).

This will happen for the lowest common multiple of 15, 25 and 100

Now, to find the lowest common multiple we can use brute force, this is, writing all the multiples of the 3 numbers and watching which one is the first to appear in the 3 lists, but here we can do another thing.

One of the numbers is 100, so its multiples are:

100, 200, 300.

And we know that with 25 we can make all of these numbers:

25*4 = 100

25*8 = 200

25*12 = 300

So all the multiples of 100 are also multiples of 25.

Then we just need to find the first common multiple between 100 and 15.

To get this, we can decompose both numbers as a product of primes:

15 = 3*5

100 = 10*10 = 2*5*2*5

Now we multiply these, but if there is some factor that appears in both numbers we remove it (only from one).

For example, here we can see that the factor 5 appears in both numbers, so we need to remove one of the fives in the product.

We will get:

(3*5)*(2*5*2) = 300

So the lowest common multiple is 300.

This means that the 300th caller will get the 3 prizes.

If you want to learn more, you can read:

link

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!