Melody

This question is definitely not answered  

b) How many positive divisors of 8400 have at least 4 positive divisors?

$${factor}{\left({\mathtt{8\,400}}\right)} = {{\mathtt{2}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{7}}$$

there are 8 prime factors here. (including duplicates)

Any one of these by itself will only have 2 factors.

The product of 2 identical ones will have only 2 factors (As Zac has already explained)

The product of any other combination will have 4 or more factors.

So products of 2 different primes:  4C2 = 6

Product of 3 primes:  222,223,225,227,557,553,552,4C3 = 11

Product of 4 primes:  2222,2223,2225,2227,2255,2235,2237,2257,5523,5527,5537,2357 =12

Product of 5 primes:  22223,22225,22227,22235,22237,22255,22257,22355,22357,23557=10

Product of 6 primes:  222235,222237,222255,222257,222355,222357,222557,223557 = 8

Product of 7 primes:  2222355,2222357,2222557,2223557 = 4

Product of 8 primes:  22223557 = 1

Total =

$${\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{11}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{52}}$$            

That is assuming I didn't miss (or double count) any      

And YES there must be a better way of counting these but I don't know what it is.  

24% of 36

I think of it like this.

% is just a rearrangment of 100 and the / (sideways 1) tells you that it is divide.

so 24% means 24/100

'of'  always means multiply

so this problems becomes

24/100*36

$${\frac{{\mathtt{24}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{36}} = {\frac{{\mathtt{216}}}{{\mathtt{25}}}} = {\mathtt{8.64}}$$

Good explanation Zac but 2-5=-3  not -7 so it is not going to be quite right.    

Doesn't really matter - your explanation of how to go about finding the answer is very good :)    

Excellent work anonymous4338 

Here is another way

$${factor}{\left({\mathtt{600}}\right)} = {{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}}$$

$${factor}{\left({\mathtt{108}}\right)} = {{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{3}}}$$

Include all factors but if the factor appears in both only include it once.

so lowest common multiple will be   $${{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}} = {\mathtt{5\,400}}$$

 I like that pic Awsomeee    LOL

I need to practice diophantine equations too.  

Thanks for providing us with one Mellie.

You always provide us with great questions. :)

Although of course we are asked to use the Euclidian algorithm (whatever that may be)

If I get around to it I will look it up later :)

Keno, which is a popular gambling game in Australian clubs, works on a similar principal to this.

I assume similar games are popular in many parts of the world but I don't know if keno is a world wide name :/

Hi Einstein Jr.

Here is how it is done    

The number of ways to choose 10 b***s from 90,  6 are drawn

So there are 84 Lose numbers and 6 Win numbers

It is easier if the winning numbers numbers are drawn first and and kept secret.

Then the player has to draw 10 ball numbers to match the winning ones (hopefully).

There are 90 numbers that can be drawn from so there are 90C10 possible draws

a) How many ways can just one be correct?

you want 9 fails and 1 success = 86C9*6C1

P(1 success)=84C9*6C1/90C10

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{9}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{9}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{6}}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.386\: \!113\: \!895\: \!203\: \!522\: \!6}}$$

b) P(2 successes)

84C8*6C2/90C10

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{8}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{8}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.114\: \!310\: \!034\: \!764\: \!200\: \!8}}$$

c) P(3 successes)

84C7*6C3/90C10

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.015\: \!835\: \!156\: \!330\: \!971\: \!5}}$$

c) P(3 successes)

84C7*6C3/90C10

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.015\: \!835\: \!156\: \!330\: \!971\: \!5}}$$

d) P(4 successes)

84C6*6C4/90C10

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.001\: \!065\: \!827\: \!829\: \!969\: \!2}}$$

e) P(5 successes)

84C5*6C5/90C10

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.000\: \!032\: \!379\: \!579\: \!644\: \!6}}$$

f) P(6 successes)

84C4*6C6/90C10

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.000\: \!000\: \!337\: \!287\: \!288}}$$

NOW for good measure

P(no successes)

84C10/90C10

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.482\: \!642\: \!369\: \!004\: \!403\: \!3}}$$

Check

P(of any number of wins)

nCr(84,9)*6/nCr(90,10)+

nCr(84,8)*nCr(6,2)/nCr(90,10)+

nCr(84,7)*nCr(6,3)/nCr(90,10)+

nCr(84,6)*nCr(6,4)/nCr(90,10)+

nCr(84,5)*nCr(6,5)/nCr(90,10)+

nCr(84,4)*nCr(6,6)/nCr(90,10)+

nCr(84,10)/nCr(90,10)

$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{9}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{9}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{6}}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{8}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{8}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{1}}$$

If you have a probelm with a particular formula, ask about that specific one.  

People often come up with some great answers :)