A pack contains n cards numbered from 1 to n. | vedclass

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(A) Let the removed cards be $k$ and $k+1$.The sum of all cards from $1$ to $n$ is $\frac{n(n+1)}{2}$.Given that the sum of the remaining cards is $12

Vedclass · Accepted Answer

$5$

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(A) Let the removed cards be $k$ and $k+1$.The sum of all cards from $1$ to $n$ is $\frac{n(n+1)}{2}$.Given that the sum of the remaining cards is $1224$,we have:$\frac{n(n+1)}{2} - (k + k + 1) = 1224$$n^2 + n - 2(2k + 1) = 2448$$n^2 + n - 4k - 2 = 2448$$n^2 + n - 2450 = 4k$$(n + 50)(n - 49) = 4k$Since $k$ is a positive integer,$(n+50)(n-49)$ must be a multiple of $4$ and $k For $n = 50$:$(50 + 50)(50 - 49) = 100 	imes 1 = 100 = 4k \Rightarrow k = 25$.Since $k Then $k - 20 = 25 - 20 = 5$.For $n = 51$:$(51 + 50)(51 - 49) = 101 	imes 2 = 202$,which is not divisible by $4$.For $n = 52$:$(52 + 50)(52 - 49) = 102 	imes 3 = 306$,which is not divisible by $4$.For $n = 53$:$(53 + 50)(53 - 49) = 103 	imes 4 = 412 = 4k \Rightarrow k = 103$.Since $k > n$ $(103 > 53)$,this is not possible.Thus,the only valid solution is $k = 25$,so $k - 20 = 5$.

$A$ pack contains $n$ cards numbered from $1$ to $n$. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is $1224$. If the smaller of the numbers on the removed cards is $k$,then $k - 20 =$

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