Lost in Addition

For the general problem: The sum of the integers from $1$ to $n$ is equal to $\frac{n(n + 1)}{2}$ (as it is the n th triangle number). Therefore, the sum of integers from $1$ to $49$ = $\frac{49 \times 50}{2}$ = 1225. Let us remove the number $x$ from the sum. This means that, excluding $x$ , the sum of the numbers to the left of $x$ is the sum of the numbers from $1$ to $x - 1$ = $\frac{(x - 1)(x)}{2}$ . This is equal to half of the sum of numbers from $1$ to $49$ (excluding $x$ ). Putting this information gives us the equation which we can then simplify:

$\frac{1225 - x}{2}$ = $\frac{(x - 1)(x)}{2}$

$1225 - x$ = $x^2 - x$

$x$ = $\sqrt{1225}$

$x$ = $35$

Therefore, the answer is $\boxed{35}$ .

For the bonus problem: This time, we need the sum of integers from $k$ to $n$ rather than from $1$ to $n$ . This can be done by getting the sum of integers from $1$ to $n$ and subtracting from this the sum of integers from $1$ to $k - 1$ . Therefore, the sum of integers from $k$ to $n$ = $\frac{n(n + 1)}{2}$ $-$ $\frac{(k - 1)(k)}{2}$ . Let us again call the number we need to find $x$ . This means that, excluding $x$ , the sum of the numbers to the left of $x$ (from $k$ to $x - 1$ ) is the sum of the numbers from $1$ to $x - 1$ minus the sum of numbers from $1$ to $k - 1$ . This is equal to $\frac{(x - 1)(x)}{2}$ $-$ $\frac{(k - 1)(k)}{2}$ . This value plus $x$ plus this value again is equal to the sum of numbers from $k$ to $n$ as the totals on either side of $x$ are equal. Therefore:

$\frac{n(n + 1)}{2}$ $-$ $\frac{(k - 1)(k)}{2}$ = $2 \times$ ${[}$ $\frac{(x - 1)(x)}{2}$ $-$ $\frac{(k - 1)(k)}{2}$ ${]}$ $+$ $x$

$\frac{n(n + 1)}{2}$ $-$ $\frac{(k - 1)(k)}{2}$ = $x(x - 1)$ $-$ $k(k - 1)$ $+$ $x$

$\frac{n(n + 1)}{2}$ $-$ $\frac{(k - 1)(k)}{2}$ = $x^2$ $-$ $k^2$ $+$ $k$

$n^2 + n$ $-$ $k^2 + k$ = $2x^2$ $-$ $2k^2$ $+$ $2k$

$n^2 + n$ $+$ $k^2 - k$ = $2x^2$

$x^2$ = $\frac{n^2 + n + k^2 - k}{2}$

$x$ = the square root of: $\frac{n^2 + n + k^2 - k}{2}$ Note: x has to be positive

Of course, $x$ has to be an integer as it is from a list of integers (as do $n$ and $k$ ). But this is the way to find the number chosen from a list with definitive start and end points.

This article I read on Brilliant site comes to my mind:

https://brilliant.org/problems/ramanujan-and-mahalanobis/ and this link also narrates the story

https://totient.wordpress.com/2008/05/28/ramanujans-continued-fraction/

Can anybody come up with the Continued Fraction which Ramanujam dictated to Mahalanobis while stirring the vegetables.