TITLE: How does this person solve the Putnam problem?
QUESTION [0 upvotes]: Consider this: 2003 A1 Putnam Solution.
I am only looking at A1 for Putnam 2003.
The problem is here: Problem A1 2003 
I would like to proceed step-by-step:   
I understand $ka_1 = a_1 + a_1 + ... + a_1$ step easily. 
I also see why $ka_1 < n$ because $a_1 + a_1 + a_1 + ... < a_1 + a_2 + a_3 + ...$
I also see:
$a_1 + a_2 + a_3 + .. < a_1 + (a_1 + 1) + (a_1 + 1) + (a_1 + 1)$
Where all $\sum a_1 = ka_1$. but how do they get: the $k - 1$ part there?
Thanks!

REPLY [2 votes]: Where they write
$$
a_1 + (a_1 + 1) + (a_1 + 1) + \cdots + (a_1 + 1) = ka_1 + k - 1
$$
is that what you're wondering about? Because $a_1$ is written $k$ times, and $1$ is written $k-1$ times (after each of the $k$ instances of $a_1$, except the first one), so that's what it becomes. Clearer ways of writing the right-hand side in that respect include $k(a_1 + 1) - 1$ and $k\cdot a_1 + (k-1)\cdot 1$.
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