1
$\begingroup$

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 $n = 5x + 7y$.

  • 0
    No, he stated it correctly. (The smallest is trivially $1$.)2017-01-31
  • 0
    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.2017-01-31
  • 2
    The 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#21189232017-01-31
  • 1
    Possible duplicate of [Largest integer that can't be represented as a non-negative linear combination of $m, n = mn - m - n$? Why?](http://math.stackexchange.com/questions/66963/largest-integer-that-cant-be-represented-as-a-non-negative-linear-combination-o)2017-01-31
  • 0
    @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.2017-01-31
  • 1
    Oops. I misread old johns comment. He basically said the same thing I did. Sorry.2017-01-31

2 Answers 2

0

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: $$g(a,b)=(a-1)(b-1)-1,gcd(a,b)=1$$ $$g(5,7)=(5-1)(7-1)-1=23$$

  • 0
    My bad, gonna fix it.2017-01-31
3

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 $(c_1c_2-c_1-c_2)$

  • 0
    By now this Q must be a duplicate of a duplicate of a ......2017-02-01