Math Questions Thread(Mathcels)

L Lawliet

Faith,Justice,Hope
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Joined
Jan 7, 2018
Messages
1,003
#1
Question : x<y<z and x,y,z are natural numbers. x+2y+3z=74 ---> what is x+y+z max value? answer:36

How can i solve that?
 

ShadowMonk

Ogreheart
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Joined
Aug 25, 2018
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1,569
#12
I'd make a permutation table in google sheets. You could also plot it as a 3d surface in some graph program/3d engine. It usually gets easier when you visualize your data.
 

Noixa

gaymaxxing with niggers
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Joined
Dec 15, 2018
Messages
356
#20
Fuck education. Only motivator to succeed in school for bluepillers it to betabux whore, for incels there is none. Just get a job as a cashier or something if you want one
 

wizardcel

sexually frustrated man. 29 yo, never had a gf
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Joined
Feb 15, 2018
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1,001
#21
Let's guess a value for Z and X

If x=1 and z=0, therefore;

1+2y=74
y=36.5 ( it's a rational number, so this can't be our answer)

if x=2 and z=0, then

2+2y=74
y=36 ( ok, it's a natural number. But let's keep going)

if x=3 and z=0
3+2y=74
y=35,5 ( again a rational number)

if x=4 and z=0
4+2y=74
y=35 ( it's a natural number, but 35<36)

Note that for every natural number greater than 1, the value of Y will decrease. Therefore, the maximum value for Y is 36. Plugging the value in x+2y+3z=74, we have 2+2x36+0=74.
 

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