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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?

 Jul 26, 2019
 #1
avatar+791 
+2

there is exactly 413 students

21*19+14 = 413

23*17+22 = 413

 Jul 26, 2019
 #2
avatar+105634 
0

Please show us some working or thought processes trvisio.

Melody  Jul 26, 2019
 #3
avatar+1675 
+1

Tip:

It's Traviso

Don't forget the A!

tommarvoloriddle  Jul 26, 2019
 #10
avatar+791 
+3

sorry...

okay melody i will try to do this next time

although i have forgoten what i had done

travisio  Jul 27, 2019
 #5
avatar+105634 
+3

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.

 Jul 26, 2019
 #6
avatar+104836 
+4

Let m be the number of columns in the first case  and  n be the  number of columns in the second

 

So....the total number of students  = the total number of students

 

23 * m   + 22    =  21 * n + 14         subtract  22 from each side

 

23m  =   21n - 8       divide both sides by 23

 

m =  (21n) / 23   -  8/23

 

We are looking for a value  of  n  such that when  21 n  is divided by 23.....the remainder  = 8

 

So we have that

 

(21n) mod (23)   =  8

 

This will happen when n  = 19  ⇒  m = 17

 

So.....the total number of  students  =  21 (19) + 14  = 413

 

 

cool cool cool

 Jul 26, 2019
 #7
avatar+105634 
0

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

 Jul 26, 2019
edited by Guest  Jul 26, 2019
 #8
avatar+104836 
+2

Yep....trial  and error......maybe not the  best  way...but   I knew that  "n" could not exceed  23  because

 

21n + 14   < 500

 

21n < 486

 

 n < 23.......

 

And since

 

(21) (23)  mod 23    =  0

 

Then I only  need to test  a few integer values < 23    before  I hit on n = 19   as the  correct solution 

 

 

 

cool cool cool

CPhill  Jul 26, 2019
 #9
avatar+105634 
0

ok thanks Chris.

Melody  Jul 27, 2019
 #11
avatar+28182 
+2

I looked at it this way:

 

N = 23m + 22        (1)                  m is an integer

N = 21n + 14         (2)                  n is an integer

 

From (1) and (2) we have:   

21n - 23m = 8

or  21(n-m) -2m = 8 

or 21(n-m)/2  - m = 4    (3)

 

Now (n-m)/2 must be an integer to satisfy (3).

 

Try (n - m)/2 = 1    (4)

Put this into (3) to get 21*1 - m = 4  or m = 17    (5)

 

From (4) and (5)   n = 2 + m = 2 + 17 = 19  (6)

Put (5)  into (1) (or (6) into (2)):  N = 23*17 + 22 = 413.

 

What if (n - m)/2 = 2 

21*2 - m = 4 so m = 38.  This time n = 4 + m = 42.  Hence N = 23*38 + 22 = 896 which is greater than 500.

 

Clearly, larger values of (n - m)/2 will produce even larger values of N, so N = 413.

 Jul 27, 2019
edited by Alan  Jul 27, 2019
 #12
avatar+105634 
0

Alan, I like your method.  It is much more simple than mine. 

 

 

I'd like to try and use it for other similar problems to make sure I fully understand.  

I think I do but I am not completely sure.

 

I think...

n-m = any positive even number     will give an integer solution for n and m, it is just a matter of chosing the correct solution given the other constraints of our question.

 

Yes I think i have it!   Thanks!

I'd still like to try doing more problems like this your way!

Melody  Jul 28, 2019
 #13
avatar+28182 
0

Thanks Melody.  This approach still requires some trial and error of course.  In this case that worked out ok at the first guess.  It won't necessarily do so for every problem!

Alan  Jul 28, 2019
 #14
avatar+105634 
0

I'd like to try it with other similar questions.

It looks like an easy way that would not normally require a lot of trial and error, just a bit I think.

Melody  Jul 28, 2019

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