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 =

We have\ 1+2+3+ * s+(n-2) ((n-2)(n-1) / 2) = 0 we get 4945. Hence n=50 n is of the form (4 p+1) or (4 p+2) etc.

A pack contains n cards numbered from 1 to n. Two cons... | (Mathematics)

We have

$\\ \begin{aligned} & 1+2+3+\cdots+(n-2) \leq 1224 \leq 3+4+ \cdots+n \\ & \Rightarrow \frac{(n-2)(n-1)}{2} \leq 1224 \leq \frac{(n-2)}{2}(n+3)\end{aligned}$ $\\$ Thus $n^2-3 n-2446 \leq 0$ and $n^2+n-2454 \geq 0$ we get $4945$ . Hence $n=50$ $n$ is of the form $(4 p+1)$ or $(4 p+2)$ etc.

Explanation

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