A and b are positive integers and a+b=2009!

what is the amount of ordered pairs of (a, b)?

Let's consider a much smaller value and work our way up to the factorial 2009!

Start with a+b = 2. The value 2 is the smallest right hand side possible since a > 0 and b > 0, so a = b = 1 is the smallest a,b possible.

The only solution to a+b = 2 is a = b = 1 as mentioned.

Now move onto a+b = 3. We have two solutions:

a = 1, b = 2

a = 2, b = 1

Now move onto a+b = 4. We have three solutions

a = 1, b = 3

a = 2, b = 2

a = 3, b = 1

Note how 'a' counts up while b counts down. I have 'a' starting at the smallest value, and b is set to add with 'a', getting to the right hand side.

So 'a' counts up to 3, while b counts down from 3.

We cannot reach 4 since 0 is not allowed for a or b.

The conjecture is that a+b = k has k-1 positive integer solutions. A simple proof of this is to list out all solutions like so

a = 1, b = k-1

a = 2, b = k-2

a = 3, b = k-3

a = k-3, b = 3

a = k-2, b = 2

a = k-1, b = 1

Where k > 2

Going from 1 to k-1 is exactly k-1 items. It's similar to how {1,2,3,..,m} has m different integers in it. Replace m with k-1 and you have the same idea.

This confirms there are k-1 solutions listed above.

The last step is to simply replace k with 2009! and we have 2009! - 1 different ordered pair solutions

Since the factorial 2009! is so massive, it's best to not expand this out using a calculator. Many calculators would produce "overflow", or similar, if you tried to compute the factorial 2009!

28

step-by-step explanation:

because that's the answer

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: