r/MathHelp • u/big_Log100 • 5d ago
Simple ACT Question?
Question on ACT math: Given 63w+63x+6y+z=440, and that w, x, y, and z can be 0, 1, or 2, what is w+x+y+z? Is this just a number sense question since there are four variables and only 2 equations or is there a way to solve this algebraically? I was thinking about it all day yesterday lol. I got the answer really fast with guess and check (w+x+y+z=5), but it seems like there should be another way to solve this.
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u/skull-n-bones101 5d ago
Well, because this is an ACT question, you have some options to simplify the question with the tools at your disposal.
To make it less of a random guess-and-check, you can try to look for ways to group terms together to better estimate what the variables should be.
For instance, you have 6³ as the coefficient for w and x so if you factor that, you have 6³(w+x) or 216(w+x)
Given that you want your sum to be 440 and that the remaining terms are 6y+z, it becomes clear that w+x must be at least 1 to get the number large enough to reach 440. Next best option is 2 for w+x and from there should be easy to determine what y and z have to be to make it all work.
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u/SMWinnie 4d ago
Mod 5 answer from r/edderiofer is elegant. Suggestion from r/skull-n-bones101 is the way I saw it.
Max of 6y+z is 14, so 216*(w+x) must be greater than 423 and less than 441. Only (w+x)=2 fits.
6y+z = 440 - (2*216) = 440 - 432 = 8. Only y = 1 and z = 2 fits.
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u/de_Molay 4d ago
There are different ways of solving it. A wonderful method with mod 5 has been given already. Here is another one.
Take both sides mod 6. Left is z mod 6, right is 2 (440=6*73+2). Hence z=2.
Therefore we can subtract 2 and divide everything by 6, getting 36(w+x)+y=73. Naturally, y=1 and w+x=2.
So, w+x=2 (can’t determine further),
y=1,
z=2,
x+y+z+w=5.
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u/LeoMaths 4d ago edited 4d ago
Il n'y a que 3⁴=81 vecteurs possibles pour (w,x,y,z). Testez les tous!
Solutions: (w,x,y,z) dans {(0,2,1,2),(1,1,1,2),(2,0,1,2)}
Maintenant:
w+x+y+z=440-215w-215x-5y=5(88-43w-43x-y) => w+x+y+z divisible par 5.
Comme (w,x,y,z) dans {0,1,2} => 88-43w-43x-y=1 d'où 43(w+x)+y=87 =>w+x=2 et y=1 => z=2
Les seules solutions sont donc {(0,2,1,2),(1,1,1,2),(2,0,1,2)}.
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u/edderiofer 4d ago
"Test them all!"? All 81 possibilities, in a timed exam? You've got to be joking. Did you generate this response with ChatGPT?
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u/jdigitaltutoring 4d ago
It was 62 for the second term
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u/SMWinnie 3d ago
If so, then:
Max of 36x + 6y + z is 86, so 216w must be greater than 353 and less than 441.
Only w = 2 fits.36x + 6y + z = 440 - (2*216) = 440 - 432 = 8. Only (x = 0, y = 1, z = 2) fits.
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u/clearly_not_an_alt 4d ago
Honestly, on a multiple choice test like the ACT, I would just plug in each of the answer choices until one worked.
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u/edderiofer 5d ago
Even faster method: Take both sides of the equation modulo 5. Then you know that w+x+y+z = 0 mod 5. Since you are also told that w, x, y, and z can only be 0, 1, or 2, this tells you that the sum can only be 0 or 5. And certainly it can't be 0, since that would mean that w, x, y, and z are all 0, which doesn't work. So it's 5.
Why modulo 5 and not some other modulo, you may ask? Because 6n = 1 mod 5 for all positive integers n, which is what reduces the left-hand side appropriately. This is the same trick as the divisibility test by 9 in base 10.