[Math] A pack contains $n$ card numbered from $1$ to $n$

algebra-precalculus

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

Best Answer

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 $$101^2-99^2=2\cdot200=400=16\cdot25\;.$$

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