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

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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