JEE Sequence And Series

Let $m$ be the number removed. Then we require that

$\dfrac{\dfrac{n(n + 1)}{2} - m}{n - 1} = \dfrac{71}{2}$

$\Longrightarrow n(n + 1) - 2m = 71(n - 1)$

$\Longrightarrow n^{2} - 70n - (2m - 71) = 0$

$\Longrightarrow n = \dfrac{70 \pm \sqrt{4900 + 8m - 284}}{2} = 35 \pm \sqrt{1154 + 2m}.$

Now if we took the root with the negative sign, we would end up with $n \le (35 - \sqrt{1154 + 2}) = 1$ , which wouldn't make sense given the conditions of removal, so we must choose the root with the positive sign.

Next, for $n$ to be an integer we will then require that $1154 + 2m = a^{2}$ for some integer $a$ , (and without loss of generality we can require that $a$ be positive). Now $1154 + 2m$ is always even, so we must have $a = 2b$ for some positive integer $b.$ Thus

$1154 + 2m = (2b)^{2} \Longrightarrow m = 2b^{2} - 577.$

For $m$ to be positive we will then require that $b \ge 17.$ Now for $b = 17$ we have $m = 1$ and thus $n = 35 + 34 = 69.$ Checking these numbers, we have that

$\dfrac{69*35 - 1}{69 - 1} = \dfrac{71}{2},$ so we have found one solution.

For $b = 18$ we have $m = 71$ and $n = 35 + 36 = 71.$ Checking these numbers, we have that

$\dfrac{71*36 - 71}{71 - 1} = \dfrac{71}{2},$ and so we have a second valid solution.

For $b = 19$ we have $m = 145$ and $n = 35 + 38 = 73$ , which would not work since we must have $m \lt n.$ In fact for $b \ge 19$ we will always have $m \gt n$ , and hence we have found the only two solutions, namely $1$ and $71$ .

Thus the desired solution is $\lfloor \sqrt{1 + 71} \rfloor = \boxed{8}.$

let $x$ be the removed number. note that $1+2+3...+n=\dfrac{n(n+1)}{2}\forall n\in \mathbb{N}$ we see that the am is $\dfrac{\dfrac{n(n+1)}{2}-x}{n-1}=\dfrac{71}{2}$ $n^2-70n-2x+71=0$ we see that for n to be natural, $D=4k^2$ $70^2-4(-2x+71)=4k^2$ for some integer k $x=\dfrac{k^2}{2}-577--(1)\longrightarrow k^2>1154--(2)$ since x is positive,note that $x\leq n\longrightarrow x\leq\dfrac{70+\sqrt{D}}{2}$ $x\leq 35+k$ insert from 1, $\dfrac{k^2}{2}-577\leq k+35$ $k^2-2k-1224\leq 0\longrightarrow -34\leq k\leq 36--(3)$ intersecting 3 and 2, $-34\leq k\leq 36 ∩ k^2>1154$ $1154\leq k^2\leq 36^2$ we see that $k^2=34^2,35^2,36^2$ hence all x $z=\dfrac{34^2}{2}-577+\dfrac{36^2}{2}-577$ as the odd value wont be an integer. $z=72$ and $\lfloor\sqrt{z}\rfloor=\boxed{8}$

Let the removed no. be $\delta$

First we note that the following expression always holds

$\frac {n(n+1)}{2}-n \le \frac {n(n+1)}{2}-\delta \le \frac {n(n+1)}{2}-1$

And given $\frac {n(n+1)}{2}-\delta =\frac{71(n-1)}{2}$

$\frac {n(n+1)}{2}-n \le \frac{71(n-1)}{2} \le \frac {n(n+1)}{2}-1$

Solving the two inequalities - From $1^{st} \text{we get} n\in[1,71]$

From $2^{nd} \text{we get } n\in(-\infty,1]U[69,\infty)$

So we get n=1 or $n \in [69,71]$

Substituting these n in $\frac {n(n+1)}{2}-\delta =\frac{71(n-1)}{2}$

There are only 2 possible $I^+$ values of $\delta =1 or 71$

Hence $\lfloor\sqrt{72}\rfloor = \boxed{8}$