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 $$1 + 27 = 28$$.