Melody

Hi Chris,

Thanks for your answer :)

I am just looking at this last bit.

(21n) mod (23)   =  8       Yes I get this.

This next step, did you work that out just by trialing lots of values?

This will happen when n  = 19  ⇒  m = 17

The American Mathematics College is holding its orientation for incoming freshmen. The incoming freshman class contains fewer than 500 people. When the freshmen are told to line up in columns of 23, 22 people are in the last column. When the freshmen are told to line up in columns of 21, 14 people are in the last column. How many people are in the incoming freshman class?

N=23Y-1        and         N=21Q+14

so

\(23Y-1=21Q+14\\ 23Y-21Q=15\\ \)

First find a solution to 23Y-21Q=1   then I will multiply through by 15 to get one solution that i actually want.

I am going to use the euclidean algorithm for this and then the extended euclidean algorithm.

23= -1(-21) +2

-21= -11(2)+1

so

-21 + 11(2) = 1

-21 + 11 [ 23+1(-21) ] =1

-21 +11(-21) +11*23   = 1

12(-21)        + 11*23   = 1

23(11     )    -21 (12   )=1     

23(11*15     )    -21 (12*15   )= 15

23(165     )    -21 (180   )= 15

So one solution will be  Y=165       and  Q= 180   

But that is not the only solution and maybe it is not the solution we want

Becasue... our N value muct be between 0 and 500

So I want a general solution

23(165     )    -21 (180   )= 15

I can add 23*21K and then take it away again and the answer will still be 15.   I do it like this.

23(165 + 21K     )    -21 (180 +23K  )= 15

So the general solution is

Y=165 + 21K       and      Q=180+23K

N= 23Y-1

N= 23[165+21K]-1

N=23*165  + 23*21K -1

N= 3795   + 483K  -1

N= 3794 + 483K

You can check by using the equation, it will work out the same.

Now

0< N < 500

0 < 3794 + 483K < 500

-3794 < 483K < -3295

-7.8 < K < -6.8

K= -7 is the only solution sind all the pronumerals used including K are integers.

So 

N = 3794 +483*-7 = 413

There must be exactly 413 students.

I understood what Chris was presenting but i do not know any of it as the Reiman's sum, is a term used more in the US. idk.

I am not surprised that you did not understand what I was saying. It needs diagrams and proper explaination and then it needs a lot of time to assimilate the ideas.

No one will understand it all on first pass.  It kind of gets absorbed with use, at least it did for me. (I am still absorbing some of it )

Maybe next time you can upload a pic and we can explain what it is trying to demonstrate.

-----------------------

I should have shown you this site in the first place.  Maybe you will understand it better than what I was trying to say. 

Actually it is almost identical to what i was trying to say only it is said better. :)

CPhill's answer is drawing criticism but it is CORRECT.

Thanks Chris :)

sqrt (14) is not 7root 2

sqrt(98)= sqrt(49)*sqrt(2) = 7sqrt2

I am a bit rusty on these but this appears to be a standard 'stars and bars' problem.

You can look that up if you want, there are some good videos on it around the place.

the balls are all the same so the only difference is how many balls are in each different box.

The balls are all the same, put them in a row.  Now you are going to seperate them with 2 bars.

The ones in front of the front bar gointo box1, whe ones between the bars go in box2 and the ones after the last bar go into box three

Draw a pic to see what I am talking about.

So now you have 6 stars and 2 bars and you just need to know how many different pleaces those bars can be put so that will be

choose 2 from 8         

8C2 = 28 ways

---------

If the boxes had alos been the same, which they weren't then...

006      

015

024

033

114

123

222

That is all I can see,  I think there would only be 7 ways .  

Please show us some working or thought processes trvisio.

Try drawing the two triangles together.  They have 2 sides that are the same length.

Both have the same base.   WX=WZ

Both have an adjoining side that is the same length. Shown by the arc.

If i make the angle between them bigger, what happens to the third side?

Three questions in a row, all very similar, who do you think your are kidding?

This was the first one, I will grant you that.