Broken Consecutive Numbers

The sum of an arithmetic series can be found using the formula, $S_n = \frac{n}{2}\left[2a+(n-1)d\right]$ Hence, the sum of $30$ consecutive integers with starting term $a$ can be written as, $S_{30} = \frac{30}{2}\left[2a + 29 \right] = 30a + 435$ Let the number that was removed be , $n = a+x, \quad 0\leq x \leq 29$ Hence we have, $S_{30} - n = 29a -x + 435 = 9909 \Rightarrow 29a = x + 9474$ Therefore, since $a$ is an integer, $29|x+9474$ . The only valid solution for $x$ is therefore $9$ in the given domain. Thus, solving for $a$ we find that $a = 327$ . Hence, we have, $n = a + x = 327 + 9 = \fbox{336}$

Let $N, N+1 ,...,N+29$ the sequence of the thirty numbers.

We have: $\sum_{i=0}^{29} N+i -(N+j)$ $= 9909$ where $N+j$ is the answer of the problem.

So $9909 = 30N +$ $\frac{29*30}{2}$ $- N -j$ so $29N - j = 9474.$ * (#) *

Don't forget that $N+j$ is a number of the sequence then $0 <= j <= 29$ that implies $29(N-1) <= 29N-j <= 29N$ and so $N=327$ .

Sobstitute N in the expression (#) and have $j = 9$ . So the answer is $\boxed{336}$

Sum of 30 consecutive numbers is $S=15(2n+29)$ . n is the first number of sequence. Let the removed number be x. $S=9909+x$ . $S>9909$ , so $n>315.8$ . Therefore x is greater than or equal to 316. $S=9909+x \ge 10225$ . Therefor $n \ge 326$ . Let n be 327 . Therefor $S=10245. x=10245-9909=336$ .

The thirty consecutive natural numbers are: $327, 328, 329, ......, 355, 356$

$327 + 328 + 329 + ..... + 355 + 356 = 10245$

If we remove $336$ from the series, then:

$(327 + 328 + 329 + ..... + 355 + 356) - 336 = 10245 - 336$ $= 9909$

Thus, the removed integer is $\boxed{336}$ .

FIRST STEP : = 9909/29 = 341.68. As 341 is average of 29 terms then its middle term means 15th term will be 341 . SECOND STEP : = 340 - 13 (340 is 14th term after subtracting we will get first term) THIRD STEP : = sum of all consecutive numbers = 326*30+sum of thirty terms = 9780 + 465 = 10245 FOURTH STEP : = 10245 -9909 (sum of 30 terms - sum of 29 terms = missing term ) = 336