Large, Really?

It follows directly from the Chicken McNugget Theorem ..........

One solution to the equation is $5(N)+4(-N)=N$ . The problem is that the coefficients are not non-negative. So, the General solution is considered. Solutions, in General, are pairs $(a,b)=(N+4t,N-5t)$ . We Need the following

$N+4t \geq 0 \implies t \geq -\frac{N}{4}$

$-N-5t \geq 0 \implies t \leq -\frac{N}{5}$

so

$-\frac{N}{4} \leq t \leq -\frac{N}{5} \ *$

if we find the Minimum $N$ (call it $N^*$ ), for which the difference $-\frac{N}{5}+\frac{N}{4}=\frac{N}{20}$ is greater than $1$ , then there surely exists an integer $t$ that satisfies (*). for all $N \geq N^*$ greater . So, for $N>20$ , we can surely write $5a+4b=N$ , for non-negative $a,b$ . We just Need to come down from $19$ and check whether the number is expressible as $5a+4b$ or not. the first number that does not have the property is $11$ .

Cheers.