# Problem of the Week Problem C and Solution A Grand Sum

## Problem

Did you know that $$1000$$ can be written as the sum of the $$5$$ consecutive positive integers beginning with $$198$$? That is, $1000 = 198 + 199 + 200 + 201 + 202$

Also, $$1000$$ can be written as the sum of $$16$$ consecutive positive integers beginning with $$55$$. That is, $1000 = 55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70$

It is also possible to write $$1000$$ as a sum of $$25$$ consecutive positive integers. This is the maximum number of consecutive positive integers that could be used to create the sum. Determine the smallest of the positive integers in this sum.

## Solution

Solution 1

Let $$n, n+1, n+2,\ldots, n+23$$, and $$n+24$$ represent the $$25$$ consecutive positive integers. Then, \begin{aligned} n+n+1+n+2+ \cdots +n+23+n+24&=1000\\ 25n+(1+2+3+ \cdots +23+24)&=1000\\ 25n+300&=1000 \\25n&=700\\ n&=28\end{aligned} Therefore, the smallest integer in the sum is $$28$$.

Note: A useful fact that we can use is that the sum of the first $$n$$ natural numbers can be calculated using the formula $$\dfrac{n(n+1)}{2}$$.

Using the formula with $$n=24$$, the sum $$1 + 2 + 3 + \cdots + 23 + 24$$ in the equation above can be quickly calculated as $$\dfrac{24\times 25}{2}=300$$.

Solution 2

Let $$n$$ represent the middle integer of the $$25$$ consecutive positive integers. Then there are $$12$$ integers smaller than the middle integer, with the smallest integer being $$(n-12)$$, and $$12$$ integers larger than the middle integer, with the largest integer being $$(n+12)$$.

Then, the sum of the $$25$$ consecutive positive integers can be written as $(n-12)+(n-11)+ \cdots+(n-1)+n +\ (n+1)+\cdots+(n+11)+(n+12)$ This simplifies to $$25n$$, because for each positive integer $$1$$ to $$12$$ in the sum, the corresponding integer of opposite sign, $$-1$$ to $$-12$$, also appears.

Thus, we have \begin{aligned} 25n&=1000\\ n&=40\\ n-12&=28\end{aligned} Therefore, the smallest integer in the sum is $$28$$.

Solution 3

In this problem, we want to express $$1000$$ as the sum of $$25$$ consecutive positive integers. The average, $$1000\div 25=40$$, is the middle integer in this sum. Solution $$2$$ is a mathematical justification of this. There will be twelve consecutive integers above the average and twelve consecutive integers below the average. Therefore, the smallest integer in the sum is $$40-12=28$$.

Solution 4

Using the note that follows Solution $$1$$, we know that the sum of the first $$25$$ positive integers is $1 + 2 + 3 + \cdots + 24 + 25 = \dfrac{25\times 26}{2} = 325$

Now, if we add $$1$$ to each term in the sum, we get $$2 + 3 +4 + \cdots + 25 + 26 = 350$$. Notice that the total increases by $$25$$. In fact, every time we increase each term by $$1$$, the total increases by $$25$$.

The number of increases by $$25$$ needed to get from $$325$$ to $$1000$$ is $$\dfrac{1000 - 325}{25} = 27$$.

Therefore, we will need $$27$$ increases for each term, and so the smallest number in the sum is $$1 + 27 = 28$$.