P112.41.2.2019.1.1 Let $a_1,a_2,\ldots, a_m$ be a set of $m$ distinct positive even numbers and $b_1,b_2,\ldots,b_n$ be a set of $n$ distinct positive odd numbers such that

\[a_1+a_2+\cdots+a_m+b_1+b_2+\cdots+b_n=2019\]
Prove that
\[5m+12n\le 581.\]
 
 

Solution

Observe that $a_1 + a_2 + ... + a_m \ge 2 + 4 + .. + 2m = m(m+1) = m^2 + m$

Also, $b_1 + b_2 + ... + b_n \ge 1 + 3 + 5 + ... + 2n-1 = n^2$.

So, we get that $n^2 + m^2 + m \le 2019$

Now, by Cauchy, we have that

$(169)(2019) \ge 169(n^2 + m^2 + m) = 169(m^2 + n^2) + 169m = (5^2 + 12^2)(m^2 + n^2) + 169m \ge (5m+12n)^2 + 169m$

And so we have that $(5m+12n)^2 \le 169(2019-m) \implies 5m+12n \le 13 \sqrt{2019-m} \le 13 \sqrt{2019}$, which gives that $5m+12n \le 584$

Now, we just need to show that it isnt possible for $5m+12n$ to be any of $584,583,582$. For this, just find a general solution and use the fact that $m^2 + n^2 + m \le 2019$. Also, $n$ needs to be odd as otherwise the whole LHS would end up being even.

So, since $5m+12n \le 584$ and we proved that it cant be any of $582,583,584$, we must have that $5m+12n \le 581$