NMTC Practice Part 11

Algebra Level 5

Out of $n$ positive integers, not necessarily different, that add up to $2014$ , $n_1$ are $1$ , $n_2$ are $2$ , $n_3$ are $3$ , . . . . . . and $n_{2014}$ are $2014$ . Determine the maximum possible value of $n_2+2n_3+3n_4+4n_5+ \ldots+ 2013n_{2014}$ .

Part of NMTC Practice Problems .

The answer is 2013.

1 solution

Mathh Mathh
Aug 21, 2014

Obviously, we'll have $n_1=0$ if we want the sum to be maximum.

$\begin{cases}n_2+2n_3+3n_4+4n_5+\cdots+2013n_{2014}=y\\2n_2+3n_3+4n_4+5n_5+\cdots+2014n_{2014}=2014\end{cases}$

$\implies 2014-y=n_2+n_3+n_4+n_5+\cdots + n_{2014}\ge 1\iff y\le \boxed{2013}$

Since it is achievable when $n_{2014}=1, n_i=0, 1\le i\le 2013, i\in\mathbb N$ , this is our answer.

Note. There was no need to force $n_1 = 0$ . We can setup the same equations with $n_1$ in them, and draw the conclusion that we want to minimize $\sum n_i$ , which is the number of positive integers, aka $n$ .

Calvin Lin Staff - 6 years, 9 months ago

Agreed. It was a very obvious observation and so I felt like using it in my solution. It changes nothing whatsoever, though.

mathh mathh - 6 years, 9 months ago

NMTC Practice Part 11

Part of NMTC Practice Problems .

The answer is 2013.

1 solution

0 pending reports