Melody

 I think that there is something wrong with your question.

That division has a remainder .....

It would be very helpful if you gave the rest of the question too.

Hi Heureka and Chris,

You have each given a really good answer here.  

I would have done it similar to Chris but Heureka's answer is really incitful.

I am talking about the first half of Hereka's answer.  I never would have thought to do that and it is so simple!

I will admit that I do not understand the second half of what you have done though Heureka. 

How did that combinatory maths sneak in there??

Rewritten:

Let f(x) be a polynomial with integer coefficients.

Suppose there are four distinct integers p,q,r,s such that  f(p) = f(q) = f(r) = f(s) = 5.   (Can there be more??)

If t is an integer and f(t)>5,

what is the smallest possible value of f(t) ?

I'd like to see this one done too, it makes little sense to me .      

Hi Chris,

Your answer is better than mine in that I certainly did not need to get into permutations.

My logic did lead to a correct answer I think (as verified below) but it made it unnecessarily messy and there was far too much room to make a numeric error.   

However:

Your answer is not correct because you have not included all the possibilities.

I'll try again. Using your ideas.

Ways to get all 5 from the same suit = 4*13C5    (Just like Chris said)

Ways to get 4 from one suit and 1 from another = 4*13C4*  39  (you forgot this one Chris)

Ways to get 3 from one suit and 2 from another = 4*13C3*   3*13C2

Total number of ways to choose 5 cards from 52 = 52C5

nCr(52,5) = 2598960 

Ways to get no more than 2 suits = 4*13C5   +  4*13C4*  39  +  4*13C3*   3*13C2

4*nCr(13,5)+4*nCr(13,4)*3*13+4*nCr(13,3)*3*nCr(13,2) = 384384

Number of ways of NOT getting 2 suits = 52C5 - 384384

nCr(52,5)-384384 = 2214576

So 

P(drawing at least 3 different suits) = 2214576 / 2598960

2214576/2598960​ = 0.8521008403361345 = 507/595

Which is exactly the same as I got doing it with permutations !!   

WOW 

Why don't you try graphing them yourself on Desmos and see what happens :))

I suggest you study the other one. 

also, I know a lot of teachers talk about cross multiply but iI think it is a confusing way to teach it.

It is easy I guess but the logic is not very obvious.

What you must ALWAYS do is the SAME thing to each side. The equation must always balance!

the pic is showing that if you add 10 to one side then you must add 10 to the other side too,

otherwise the scale would become unbalanced!

Now ot your p[roblem:

\(y = \frac{1}{x}\\ \text{First make both sides a fraction}\\ \frac{y}{1} = \frac{1}{x}\\ \text{Now you want x on the top so turn BOTH sides upside down (take the reciprocals)}\\ \frac{1}{y}= \frac{x}{1}\\ \frac{1}{y}= x\\ \text{Now swap sides}\\ x=\frac{1}{y}\)

It is NOT    y=x/z + 1      IT IS      t=x/(z+1)      there is a big difference  

\(x = yz + y \)

the aim is to make y the subject so the first thing you have to do it to facot out the y (so that you will only see one y)

\(yz+y =y*z+y*1= y(z+1)\)

so

\(x = yz + y\\ x=y(z+1) \)

Now you want the y bt itself. At the moment it is multipled by (z+1) the opposite of multiply is divide. 

SO you have to divide BOTH sides by (z+1)

\(x = yz + y\\ x=y(z+1)\\ \frac{x}{z+1}=\frac{y(z+1)}{z+1}\\ now \;\;cancel\\ \frac{x}{z+1}=y\\ \text{Now swap sides}\\ y=\frac{x}{z+1} \)

Mike draws five cards from a standard 52-card deck. What is the probability that he draws a card from at least three of the four suits? Express your answer as a simplified fraction.

I'll give it a go :)

There are 52*51*50*49*48 =  311875200  permutationss of 5 cards altogether (with no restrictions)

I'm going to try and work out how many of these do not include at least 1 card from 3 different suits.

5 hearts

1*(13*12*11*10*9) 

4 hearts and one other card

5*(13*12*11*10*39)

3 hearts and 2 from one other suits

5C2* (13*12*11*13*12)*3  

And this can be multiplied by 4 beacuse there are 4 suits 

so

 Number of combinations without 3 different suits represented

= [1*(13*12*11*10*9)   +  5*(13*12*11*10*39)   +  5C2* (13*12*11*13*12)*3 ]    *4

=(13*12*11*4)[ 90  +  5*(390)   +  10* (156)*3  ]

= [(13*12*11*4)[ 90  +  1950   +  4680  ]

= [(13*12*11*10*4)[ 9  +  195  +  468  ]

= 13*12*11*10*4* 672 

So the number of combinations with at least 3 different suites represented 

=  52*51*50*49*48 - 13*12*11*10*4* 672 

= 12*4   [ 52*51*50*49   -   13*11*10* 672 ]

= 12*4*10*13   [ 4*51*5*49   -   11* 672 ]

= 12*4*10*13*4   [ 51*5*49   -   11* 168 ]

= 12*4*10*13*4 *7  [ 51*5*7   -   11* 24 ]

= 12*4*10*13*4 *7 *3 [ 17*5*7   -   11* 8 ]

= 3*4*4*7 *10*12*13 [ 595   -   88 ]

= 3*4*4*7 *10*12*13 *507

So the prob of not getting at least 3 suites 

\(=  \dfrac{3*4*4*7 *10*12*13 *507}{52*51*50*49*48}\\ =  \dfrac{507}{17*5*7}\\ =  \dfrac{507}{595}\\\)