Melody

Yes, I got that as well.

Thanks guest,

I have some interest.

Did you use code to do the grunt work and just test all the possibilities

Or did you actually come by the answer in a logical, mathematical (no trial and error)  way ?

Thanks for answering. But it is very hard to read and comprehend when it is all squashed up like that.

First put the left side into its simplest form.  Can you do that? 

(you can use a calc if you have to but make sure it stays in fraction form)

Watch this

The positive five-digit integers that use each of the digits 1, 2, 3, 4 and 5 exactly once are ordered from least to greatest. What is the 50th integer in the list?

Yes there is a quicker way.

How many begin with 1?   that would be 4!=24

The next 24 would begin with 2

So the first 48 begin with either 1or 2

The 49th one will be     31245

and he 50th one will be  31254

Hint 

\(f(2)-2f(\frac{1}{2})=4^2\\ and\\ f(\frac{1}{2})-2f(2)=4^{0.5}\)

solve these simultaneously to find f(2)

Please let asker guest finish this.

For how many integers n where 2≤n≤100 is (n2) odd? 

1^2=1

2^2=4

3^2=9

5^2=25

6^2=36

7^2=49

8^2=64

9^2=81

so all odd numbers will have an odd square.   How many odd numbers are there between 2 and 100?

I have assumed that n2 is n^2. 

In retrospect, that is probably not what you meant but we should not have to be mind readers.

answered.

I think it is 6

\((b+1)^4=b^4+4b^3+6b^2+4b+1 \)

The sum of the digits will always be 16 base 10 which is 2^4 for any base greater than 2

However this can only be true is 6 is allowed in the base.

The biggest base that does NOT allow the use of 6 is base 6

I have not tried it with examples, if someone wants to refute this with an example then I would like to see it. :)