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## 'Weekly Challenge 6: AP Train' printed from http://nrich.maths.org/

This problem is based on the formula for the sum of the first $n$
natural numbers

$$

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

$$

We have

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: $1111 = 11\times 101$. Thus, for
solutions we require

$$

D = 11, N=102\quad\mbox{ or } D=101, N=12

$$

You can see these two sums in action on this spreadsheet
screenshot