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if Chicken McNuggets are sold in containers of sizes $x$ and $y$ where $\gcd x,y = 1$ then it is known that the maximum number of nuggets that it is impossible to buy exactly is $xy - x - y$.

but what if Chicken McNuggets are sold in containers of sizes $x$, $y$, and $z$? with $\gcd x, y = 1$ and $\gcd x,z = 1$ and $\gcd y, z =1 $? im gonna say its probably something like $xyz - xy - xz - yz + x + y + z$ but thats probably completely wrong lol

for two nugget boxes it's the same as $(x-1)(y-1) -1$ so maybe for $3$ nugget boxes it will be $(x-1)(y-1)(z-1)-1$? so that's $xyz - xy - xz - yz + x + y + z$

terrace
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1 Answers1

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The original question is about boxes of 6, 9 and 20 nuggets. So, all we need to do is trying to write the smallest possible number as a sum in terms of 9 and 20 for each x ≡ y mod 6 where y : {0,1,2,3,4,5}.

x ≡ 0 mod 6 (we have 6, which is an exception for not using 9 nor 20.)

x ≡ 1 mod 6 (we have 49 as the smallest : 20+20+9. The next term congruent to 1 mod 6 is of course 49+6=55 and so on.)

x ≡ 2 mod 6 (we have 20 as the smallest. The next term is 20+6=26 and so on.)

x ≡ 3 mod 6 (we have 9 as the smallest. The next term is 9+6=15 and so on.)

x ≡ 4 mod 6 (we have 40 as the smallest : 20+20. The next term is 40+6=46.)

x ≡ 5 mod 6 (we have 29 as the smallest : 20+9. The next term is 29+6=35.)

The Frobenius number is 49(the largest on the list)-6 = 43

It is the easiest way of solving this problem.

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