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# Weekly Challenge 6: AP Train

##### Stage: 5 Short Challenge Level:

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