# AP train

An arithmetic progression is shifted and shortened, but its sum remains the same...

## Problem

A particular list of $N$ consecutive integers starts with $1111$ as follows:

$$1111, 1112, 1113, \dots, 1111 + N-1$$

The entire list is shifted $D$ places along the number line and the first number then excluded, leaving a list of $N-1$ larger consecutive numbers as follows:

$$

1112+D, 1113+D, \dots, 1111 + N -1+ D

$$

In each list the sum of the integers is the same.

What are the possibilities for $N$ and $D$?

Did you know ... ?

Progressions of integers occur remarkably frequently in mathematics in applications from quantum mechanics to number theory and they have many beautiful properties. Even Carl Gauss, possibly the greatest mathematician of all time, fondly recalled his first encounter with sums of consecutive natural numbers, when he noticed that the sum of the first $100$ whole numbers equalled $50$ lots of $101$.

Progressions of integers occur remarkably frequently in mathematics in applications from quantum mechanics to number theory and they have many beautiful properties. Even Carl Gauss, possibly the greatest mathematician of all time, fondly recalled his first encounter with sums of consecutive natural numbers, when he noticed that the sum of the first $100$ whole numbers equalled $50$ lots of $101$.

## Student Solutions

**Forming an equation using the amounts added and subtracted**

To shift the sequence along $D$ spaces,

from $(1111)+ (1112)+( 1113)+, \dots, (1111 + N-1)$

to $(1111+D)+(1112+D) + (1113+D) + \dots+(1111 + N -1+ D),$

you add $D$ to each term, so you add it $N$ times, since there are $N$ terms.

That is the same as adding a total of $N\times D$.

Then removing the first term, $1111+D$, takes the sum back to its original value.

So $N\times D$ must be equal to $1111+D$.

$$\begin{align}ND&=1111+D\\\Rightarrow ND-D&=1111\\\Rightarrow (N-1)D&=1111\end{align}$$

$N$ and $D$ are both integers, so $N-1$ and $D$ must be factors of $1111$.

$1111=11\times101$, which are both prime, so all of the possible values of $N$ and $D$ are shown below:

Product $(N-1)\times D =1111$ | $N$ and $D$ |
---|---|

$1111\times1$ | $N=1112,D=1$ |

$1\times1111$ | $N=2,D=1111$ |

$11\times101$ | $N=12,D=101$ |

$101\times11$ | $N=102,D=11$ |

You can see some of these sums in action on the spreadsheet screenshot at the bottom of the page.

**Using the formula for the sum of the first $n$ natural numbers**

The sum of the first $n$ natural numbers is equal to $$

1+2+\dots + (n-1) + n =\frac{1}{2}n(n+1)

$$

Its easier in a problem like this to introduce some notation. Let's write $S(n)$ to mean the sum of the first $n$ natural numbers.

Since the two sums are equal we have

$$

S(1111+N-1)-S(1110) = S(1111+N -1 +D) - S(1111 + D)

$$

Let's put the formula into each of these and cancel each factor of a half.

$$

(1110+N)(1111+N)-(1110)(1111)=(1110+N+D)(1111+N+D)-(1111+D)(1112+D)

$$

Before leaping in we can see that many parts cancel, so we can put

$$

2221N + N^2 = 2221(N+D) + (N+D)^2 - 2223D - D^2- 2222

$$

Collecting things together gives

$$

D(N-1) = 1111

$$

Now for a bit of number theory: $1111 = 11\times 101$, or $1\times1111$. Thus, for solutions we require:

$D = 11, N=102$ or $D=101, N=12$ or $D=1,N=1112$ or $D=1111,N=2$.

You can see some of these sums in action on this spreadsheet screenshot

Image