A pack contains $n$ card numbered from $1$ to $n$ A pack contains $n$ card numbered from $1$ to $n$. Two consecutive numbered cards are removed from
the pack and sum of the numbers on the remaining cards is $1224$. If the smaller of the numbers on
the removing cards is $k$, Then $k$ is.
$\bf{My\; Try}::$ Let two consecutive cards be $k$ and $k+1,$ Then given sum of the number on the 
remaining cards is $1224$ .So $\left(1+2+3+.........+n\right)-\left(k+k+1\right) = 1224$
So $\displaystyle \frac{n(n+1)}{2}-(2k+1) = 1224\Rightarrow n(n+1)-(4k+2) = 2448$
Now I did not understand how can i calculate value of $(n,k)$
Help Required
Thanks 
 A: You have $n^2+n-4k-2450=0$. Treat this as a quadratic in $n$ with constant term $-4k-2450$. Clearly we need a positive value, so
$$n=\frac{-1+\sqrt{1+16k+9800}}2=\frac{-1+\sqrt{16k+9801}}2\;,$$
$9801=99^2$; $100^2-99^2=199$, which is not a multiple of $16$, but $$101^2-99^2=2\cdot200=400=16\cdot25\;.$$
A: Your last equation becomes $(n+1/2)^2=(16k+9801)/4$.  The numerator must be a perfect square.
A: A couple of observations that may give some hints to you:


*

*You can get in the right ballpark by looking for values of $n$ that give you something a bit above $1224$ for your formula for $n$.

*Since the sum after the cards are removed is even, the sum before the cards were removed is odd.  This means that $n$ is not evenly divisible by $3$.

A: The key additional piece of information is that you know that $k$ is between $1$ and $n-1$.  Since $n(n+1)=2448+(4k+2)$, this means that $n(n+1) \geq 2448+(4\cdot1+2) = 2454$ and that $n(n+1) \leq 2448+4(n-1)+2$, or (rearranging the terms) that $n(n-3) \leq 2446$.  You can complete the squares to solve these quadratic equations and get bounds on $n$; once you do this, you'll find that there's only one value of $n$ in the appropriate range.  Once you have $n$, then you can find $n(n+1)$ easily, and from there you can solve for $k$ using your formula.
A: $(n, k) = (50, 25)$
I think you are almost there.  If I understand your question right you just solve for $k$ in terms of $n$. 
\begin{align*}
& n(n+1) - (4k + 2) = 2448 \\
\implies & n(n+1) - 2 - 2448 = 4k \\
\implies & n(n+1) - 2450 = 4k \\
\implies & \frac{n(n+1) - 2450} 4 = k
\end{align*}
So $(n, k)$ can be written $$\left(n, \frac{n(n+1) - 2450} 4\right).$$
So then you just need to find an $n$ such that both $n$ and $k$ are both positive and integers (You can't have a fraction of a card).
This next part probably goes beyond pre-calc, but looking at your profile, I think you can handle it.
We need $n(n+1) - 2450 \equiv 0 \pmod 4$. This just means that $n(n+1) - 2450$ divided by $4$ has a remainder of $0$.
Modular arithmetic is a bit different than what you may be used to in precalc, but it goes like this (if I remember correctly, its been a while).
\begin{align*}
& n(n+1) - 2450 \equiv 0 \pmod 4 \\
\implies & n(n+1) \equiv 2450 \pmod 4 \quad \text{You can replace $2450$ with the remainder of $2450$ divided by $4$.}\\
\implies& n(n+1) \equiv 2 \pmod 4
\end{align*}
There are 4 potential answers to check,  0,1,2,3 (after that they just replace such that $n + 4 \equiv n + 8 \equiv n + 12\dotsb$)
At $n = 1 \implies 1\cdot(1+1) / 4$ the remainder is $2$
At $n = 2 \implies 2\cdot(2+1) / 4 = 6/4$ the remainder is $2$
Last part.  You cannot have fewer than zero cards so $\frac{n(n+1) - 2450}  4 > 0$.  This is a simple inequality. $n > 49$.
So the answer is $\left(n, \frac{n(n+1) - 2450} 4\right)$, where $n = 1 + 4x$, $n = 2 + 4x$ where $x$ is any positive integer and $n > 49$.
After looking at Steven Stadnicki's answer, I saw there is one more restriction on $n$.  $n$ cannot be such that $n < k$.  So you can set up an inequality, 
$ n > \frac{n(n+1) - 2450} 4 \implies n \le 51$.
So given all our criteria for valid $n$ values, the only allowed n is:
$n=2 + 12 \cdot 4$
The solution for $(n, k)$ is thus $(50, 25)$.
That should work even if it's a little longer then the other answers proffered
A: Sum of 1st n natural nos. $=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
Given: $\frac{\mathrm{n}(\mathrm{n}+1)}{2}-(\mathrm{k}+\mathrm{k}+1)=1224$
$$
\Rightarrow \frac{\mathrm{n}(\mathrm{n}+1)}{2}-2 \mathrm{k}=1225 \ldots .(1)
$$
also
$\frac{\mathrm{n}(\mathrm{n}+1)}{2}-(2 \mathrm{n}-1)\left(\right.$ when $\mathrm{k}=(\mathrm{n}-1)<1224<\frac{\mathrm{n}(\mathrm{n}+1)}{2}-3($ when $\mathrm{k}=1)$
$\frac{\left(n^2-3 n+2\right)}{2}<1224<\frac{\left(n^2+n-6\right)}{2}$
$(n-2)(n-1)<1448<(n-2)(n+1)$
So $n=50$ by wt and trial method
So in equation $(1)(51)\left(\frac{50}{2}\right)-2 k=1225$
$1275-2 \mathrm{k}=1225$
$$
\mathrm{k}=25
$$
