What is the largest positive integer $n_0$ for which there are no $x, y ∈ \mathbb{Z}$ with $x, y ≥ 0$ so that $n_0 = 5x + 7y$?
Give a proof that if $n > n_0$ then there are non-negative integers $x$ and $y$ so that $n = 5x + 7y$.
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No, he stated it correctly. (The smallest is trivially $1$.) – TMM Jan 31 '17 at 23:08
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From some point onwards, every integer is representable as $5x+7y$, so it does make sense to ask for the largest which cannot be so represented. – Old John Jan 31 '17 at 23:10
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2The largest integer not possible is $(5-1)(7-1)-1=23$. see http://math.stackexchange.com/questions/2118907/guess-the-number-of-red-balls/2118923#2118923 – Emax Jan 31 '17 at 23:11
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1Possible duplicate of Largest integer that can't be represented as a non-negative linear combination of $m, n = mn - m - n$? Why? – Sarvesh Ravichandran Iyer Jan 31 '17 at 23:28
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@OldJohn. But the small integers can not be represented. 1,2,3,4 are impossible as they are all less than 5. 6, 8,9, 11 are impossible. This isn't a "how far can I go before something stops" question. This is a "how far do I have to go before something starts" question. – fleablood Jan 31 '17 at 23:55
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1Oops. I misread old johns comment. He basically said the same thing I did. Sorry. – fleablood Jan 31 '17 at 23:56
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This is the Frobenius coin problem. For given $c_1,c_2$ with $\gcd(c_1,c_2)=1$, the largest number which cannot be represented by $ac_1+bc_2$ with $a,b \ge 0$ is $(c_1c_2-c_1-c_2)$.
The intuition I prefer to understand this is to see what the value of multiples of the larger coin, say $c_1$, is modulo the smaller $c_2$, and see that the last modular equivalence is filled at $c_1c_2-c_1$, which means it must still be open at $(c_1c_2-c_1-c_2)$
Joffan
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There is a formula if the least common divisor of the 2 numbers (5, 7) is equal to 1, which is always true for primes: $$g(a,b)=(a-1)(b-1)-1,gcd(a,b)=1$$ $$g(5,7)=(5-1)(7-1)-1=23$$
Zonko
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