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Here is part 4  :)))

EVERYBODY SLAP ROSALA! (AKA ROSETHROWER) 

XD XD

OH MY GOD!what am i hearing this.,....

a letter from RoseThrower..

Dear Captain,

i did not know you were such a violent muscular robot ,dont you know the difference between a girl and a boy..my name is RoseThrower...and thats female...you will beta me up...hawww you will beta up a girl that is so bad,......you dont deserve to be the quen protector.....so bad

with SLAPS

RoseThrower

you have to admit, the queen is just very royal... and I'll take the icecream coupons!!

Here's a pic......

The area of this quadrilateral is just B x H  = (3 x 3)  = 9 sq units.......

Yep Chris, we are both brilliant     LOL   :))

Thanks Mellie for making us think.  We can always rely on you for that :)

Melody, I agree with your answer to (b).....I was assuming unique primes.....but clearly there could not be 4 unique primes that would produce a positive interger with exaxctly 20 divisors.....so .....your answer of 3 is correct.....

Sorry Chris I did not realize you had already answered - I only saw your first answer.

You must have posted the other s when Iwas off doing things and I did not realize that they were there. 

Anyway, we came to the same conclusion  :/

b)

Perhaps I have the answer - using your algebra Chris :)

The factors of 20 are 1,2,4,5,10,20

Now if we are going to have the largest number of primes to make 20 factors we need the most integers greater than 1 that can multiply to give 20.

they are 2*2*5

So the answer is 3 unique primes (6 primes altogether).  This is the maximum number of prime numbers that can divide this composite number with 20 factors.

a * b * c^4      

c) So the smallest positive integer that has exactly 20 positive divisors will be

3*5*2^4

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{4}}} = {\mathtt{240}}$$

Thanks Chris - this was a real joint effort.   Are you happy with the answer ?   

Chris, I do not see how you have answered the question.

(Although i did get something out of your algebra logic.)

You have found - like i did - that the product of 4 unique primes can give an answr with 16 factors.

BUT the question calls for EXACTLY 20 factors.  Do you think that such a number does not exist?  

c)  I believe the answer to this is 240....to see why......

As per my explanation to  (b) above, the largest numer of primes that could factor such a  number is 4.....

Note that  2,3 ,5 and 7   are the smallest primes......then...using the reasoning from (b), above.... we are looking for four exponents, that, when 1 is added to each and all are multiplied together, would equal 20.

But no such integers  k, l, m and n  exist such that (k + 1)(l + 1)(m + 1) (n + 1)  = 20  ...  where k, l ,m and n ≥ 1......so.....this number, whatever it is, can't have 4 prime factors

Let's drop 7 out of the mix and suppose it has just 3 prime factors...2, 3 and 5....again  we are looking for three exponents,  that, when 1 is added to each and all are multiplied together, would equal 20.  Put another way, we are looking for k, l and m ≥ 1   such that (k + 1)(l + 1)(m + 1) = 20

Note that the  only possibility here  is when we have  2 *2 *5  = 20.....and the smallest possible product would be 2^(4 )* 3^(1) * 5^(1) =  2^4 * 3 * 5 = 240

Now......the only remaining posibility is  that this number is composed of the two smallest primes, 2 and 3,  and we are looking for  some k  and l ≥ 1 such that (k + 1)(l + 1) = 20.....clearly, the only possibilities  are when k = 4 and l = 5, or vice-versa

So this number would factor as either 2^3 * 3^4   = 648   or 2^4 * 3^3 = 432....and both are > 240......

wow, captain man.

there was a crime proceeding and you didnt even tell me there was.

okay, today you're attacking the alligator, while today I'll protect Camelot

that alligator could've killed me yesterday

captain man, is this girl seriously trying to steal the queens throne!?

sounds like my little sister Piper forcing my friend jasper to d be her servant slave thingy majiga

another attempt to steal the Queen's throne, huh?

did not know you wanted the throne so badly. now you're behaving worse than Titanium Rome with his trash talk (which honestly needs some work).

no matter what, RoseThrower, you'll never take away the Queen's throne away and take away her title as queen. Never, Rosethrower, becuz, its called stealing, and if you do that, I will be able to beat up up and take you TO JAIL. and nobody wants to be in jail, cuz thats the 2nd worst place ever. oh, you like jail? no problem, cuz then i'll take you to the worst catagory place ever, which would be the hospital, nurse, dentist, pharmacy, etc. cuz then i can think of many ways to put you to sleep. so if you dont want to go to jail, or those other places ive mentioned, do not steal the throne (dont even try), then be a good girl, then i'll give you a free roses for your date/crush coupon, and 2 free large sized ice cream coupons

There is an app for it but it is an online tool, sorry.

b) .....I believe the answer to this is 4.....to see why......

Note that.....the number of possible divisors for a^k (where a is prime, and k ≥ 1)  is just (k + 1)

For instance, 3²  = 9, has  (2 + 1) = 3 possible divisors...... 1, 3 and 9

And the number of possible divisors for a^k * b^m    where a, b are prime, and k, m ≥ 1 .....  is just   (k + 1) (m +1)

So.....consider a positive integer factored as a^k * b^m *cⁿ * d^p  where a, b, c, d are prime and k, m, n and p  = 1

Then, the possible number of divisors is just (k + 1)(m + 1)(n + 1)(p + 1)  = (2)(2)(2)(2)  = 16

Note that, if another prime (also to the first power)  factored the number, we would have (2)(2)(2)(2)(2)  = 32 possible divisors.....and that's too  many.....!!!!

Edit...this answer is incorrect......Melody has the correct solution, below.....!!!!!

@@ End of Day Wrap  Tues 30/6/15   Sydney, Australia Time   10:45pm  ♪ ♫

Hi all,

This time our illustrious answerers were Squareroot, Alan, Mellie, CPhill, Sir-Emo-Chappington, Anonymous4338, Radix, Zacismyname, TitaniumRome, Rosala and Nauseated.  Thanks all  

$${\mathtt{0.7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.8}} = {\frac{{\mathtt{3}}}{{\mathtt{2}}}} = {\mathtt{1.5}}$$

Answers are fiven as fractions and as decimals.