NotThatSmart

In order to solve this problem, we must find the common ratio of the numbers. 

Since it's a geometric series, setting r as the common ratio, we get

\(r = .036 / .4 = 36 / 400 = 9 /100 = .09 \)

We can calculate the sum using a really handy formula. 

\(\frac{a}{1-r}\) where a is the first number and r is the common ratio. 

Plugging in the values we already now, we get that the sum is equal to

\( .4 / [ 1 -.09] = .4 / .91 = 40 /91 \)

So our final answer is 40/91

Thanks! :)

The smallest possible prime divisor a number can have is of course, \(2\)

If we check the multiplication, we can see that 3 odd numbers can't possibly mulitply to be an even. 

So, \(5^{19} * 7^{13} * 3^{31}\) is not divisble by 2. 

The next smallest is 3. 

The third term of the multiplication is \(3^{31}\)

This means, no matter what, this number has to be divisible by 3. 

So 3 is our final answer. 

Thanks! :)

Let's multiply everything out and see what we have. 

We get 

\(x^3(x^2+tx)\\ x^5+x^4t\)

As shown, there is not x^2 term. 

This means that t can pretty much be anything as long as it doesn't contain x^-2

Thanks! :)

We can convert both numbers to base 10, add them up, and convert them back. 

First, let's convert the two. We have

\((357)_8 = (3 × 8^2) + (5 × 8^1) + (7 × 8^0) = (239)_{10} \)

\((501)_8 = (5 × 8^2) + (0 × 8^1) + (1 × 8^0) = (321)_{10} \)

Adding our two results up, we get the sum

\(321+239=560\)

Now, we must convert this into base 8 again. 

We get \(560_{10}=1060_8\)

So our final answer is 1060. 

Thanks! :)

In order to solve this problem, we can use a very simple trick. 

First, let's set \(x = 0.\overline{2123}\)

If we have this for the value of x, we get \(10000 x = 2123.\overline{2123}\)

Now, we subtract x from 10000x. We get

\(10000x-x= 2123.\overline{2123} -\overline{2123}\\ 9999x=2123\)

This means we get

\(x = \frac{2123}{9999}\)

Simplifying, we get

\(x = \frac{193}{909}\)

So our final answer is \( \frac{193}{909}\)

Thanks! :)

First, let's put s in terms of t so we can do some subsitutions. We get

\(\quad s=-\frac{2\left(-4t+1\right)}{5}\) from the first equation. 

Subsituting this value of s into the second equation, we get

\(3t-6\left(-\frac{2\left(-4t+1\right)}{5}\right)=9-2t\)

Now, we solve for t. We get

\(\frac{-23t+12}{5}=9\)

\(-23t+12=45\)

\(t=-\frac{33}{23}\)

Now, we subsitute t back into the first equation to find s. We get

\(s=-\frac{62}{23}\)

We get \(s=-\frac{62}{23},\:t=-\frac{33}{23}\)

So our final answer is \((-62/23, -33/23)\)

Thanks! :)

We don't have to consider any numbers ending with 5 or even, since we know they are composite numbers. 

,

So now, we go through a list. 

First, start with 2001. We have \(2001=3*667\), and it has 8 factors, so it is composite. 

2003 is prime, as it is only divisible by 1 and itself. 

We have \(2007=3*669\), so 2007 is composite. 

\(2009 = 7*287\), so 2009 is composite. 

\(2011\) is prime. 

\(2013=671*3\), so 2013 is composite.

\(2017 \text{and} 2027\) are prime, but everything in between is composite. 

\(2029\) is prime, so that's our final answer. 

So 2029 is our answer. 

Thanks! :)

A prime number is a number only divisble by 1 and itself. 

Let's take a look at the numbers we have. 

First, we have \(1\). It is only divisble by 1 and itself, but it only has one true factor since \(1=1*1\), so it is not consdered prime. 

Next, we have 2. It is divisble as \(2 = 2*1\), so it is prime. 

Going through the list, \(3= (3*1), 5 = (5*1), 7=(1*7)\) are all prime numbers. 

However, \(4 = (2*2), 6 = (2*3), 8 = (2*4), 9=(3*3), 10=(2*5)\)are all composite numbers. 

So our final answer is \(2,3,5,7\)

Thanks! :)

First, let's subtract the to numbers and see what we get. 

We get \(123 - 47 = 76\)

Now, we need to find the prime factors of 76. 

When we factorize it, we get \(76 = 2^2 \cdot 19\)

This means there are only two distinct prime factors. \(2, 19\)

Adding these two up, we get \(19+2=21\)

So 21 is our final answer!

Thanks! :)