Hard Word Problem

Apr 2015
684
7
Bronx, NY
A man invited 100 persons in a party, they included Engineers, Scientists and Workers. He had 100 plates for the dinner. Now Engineers saw it and thought why don't we play our game. They said we will need two plates per person for the dinner. Seeing that Scientists too couldn't help, they said we will need 4 plates per person for the dinner. Now workers thought, they will fail the dinner. So they said we need only 1 plate per four person. At last the dinner was successful all 100 plates were used by the 100 persons. So can you tell me the respective number of Engineers, Scientists and Workers?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Looks very straight forward to me. Let x be the number of engineers. Allowing two plates per engineer, that requires 2x plates. Let y be the number of scientists. Allowing four plates per scientists, that requires 4y plates. Let z be the number of workers. Allowing one plate for every four workers, that requires z/4 plates.

So we need 2x+ 4y+ z/4= 100 with the understanding that x, y, and z are integers.

Multiplying through by 4, 8x+ 16y+ z= 100. I would also write that as 8x+ 16z= 100- z. Since the left side is obviously divisible by 8, so is 100- z. Write 100- z= 8a and divide the equation by 8: x+ 2y= a.

That is a Diophantine equation (maybe not all that "straight forward"). In particular, taking x= -1, y= 1 we have x+ 2y= -1+ 2= 1. Multiplying by a, if x= -a, y= 2a, x+ 2y= -a+ 2a= a.
Further, it is easy to see that x= -a+ 2k, y= a- k is also a solution for any integer k: x+ 2y= (-a+ 2k)+ (2a-2k)= a.

But a was equal to 100- z so we have x= z- 100- k and y= 100- z+ k. Those are again Diophantine equations: z- x= 100+ k and y+ z= 100- k. Solve those for x and z. Put them back into the original equation to determine k.
 
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Feb 2015
2,255
510
Ottawa Ontario
A man invited 100 persons in a party:
they included Engineers, Scientists and Workers.

He had 100 plates for the dinner.

Now Engineers saw it and thought why don't we play our game.
They said we will need two plates per person for the dinner.

Seeing that Scientists too couldn't help,
they said we will need 4 plates per person for the dinner.

Now Workers thought, they will fail the dinner.
So they said we need only 1 plate per four person.

At last the dinner was successful all 100 plates were used by the 100 persons.
So can you tell me the respective number of Engineers, Scientists and Workers?
................................................
2 solutions:

E = 15, S = 13, W = 72

E = 30, S = 6, W = 64

To have a unique solution, you need another information,
like "2 more E's than S's".
 
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Matt Westwood

MHF Hall of Honor
Jul 2008
1,281
426
Reading, UK
Looks very straight forward to me. Let x be the number of engineers. Allowing two plates per engineer, that requires 2x plates. Let y be the number of scientists. Allowing four plates per scientists, that requires 4y plates. Let z be the number of workers. Allowing one plate for every four workers, that requires z/4 plates.

So we need 2x+ 4y+ z/4= 100 with the understanding that x, y, and z are integers.

Multiplying through by 4, 8x+ 16y+ z= 100. I would also write that as 8x+ 16z= 100- z. Since the left side is obviously divisible by 8, so is 100- z. Write 100- z= 8a and divide the equation by 8: x+ 2y= a.

That is a Diophantine equation (maybe not all that "straight forward"). In particular, taking x= -1, y= 1 we have x+ 2y= -1+ 2= 1. Multiplying by a, if x= -a, y= 2a, x+ 2y= -a+ 2a= a.
Further, it is easy to see that x= -a+ 2k, y= a- k is also a solution for any integer k: x+ 2y= (-a+ 2k)+ (2a-2k)= a.

But a was equal to 100- z so we have x= z- 100- k and y= 100- z+ k. Those are again Diophantine equations: z- x= 100+ k and y+ z= 100- k. Solve those for x and z. Put them back into the original equation to determine k.
I'm puzzled. I can see:
2x+ 4y+ z/4= 100

but multiplying thru by 4 I get:
8x+ 16y+ z= 400

Is there something I've missed?
 
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