Calculate Number Offset for Removed Cards from Sequence

To find the value of $k - 20$, we set up the equation for the sum of the numbers on the remaining cards.

Let the total number of cards in the deck be $n$. The sum of all numbers from $1$ to $n$ is given by the sum of an arithmetic progression: $S_n = \frac{n(n+1)}{2}$

Two consecutive cards labeled $k$ and $k+1$ (where $1 \le k \le n-1$) are removed. The sum of these two removed cards is: $\text{Sum of removed cards} = k + (k + 1) = 2k + 1$

According to the problem, the sum of the remaining cards is $1224$: $\frac{n(n+1)}{2} - (2k + 1) = 1224$

Rearranging this equation, we get: $\frac{n(n+1)}{2} - 2k = 1225$ $\implies 2k = \frac{n(n+1)}{2} - 1225$

Since $1 \le k \le n - 1$, we have the inequality bounds for $2k$: $2 \le 2k \le 2n - 2$

Substituting $2k = \frac{n(n+1)}{2} - 1225$ into the inequality: $2 \le \frac{n(n+1)}{2} - 1225 \le 2n - 2$

Lower Bound for $n$:

$\frac{n(n+1)}{2} - 1225 \ge 2$ $n(n+1) \ge 2454$

Testing integer values for $n$:

• For $n = 49$: $49 \times 50 = 2450 < 2454$
• For $n = 50$: $50 \times 51 = 2550 \ge 2454$

So, $n \ge 50$.

Upper Bound for $n$:

$\frac{n(n+1)}{2} - 1225 \le 2n - 2$ $n(n+1) - 2450 \le 4n - 4$ $n^2 - 3n - 2446 \le 0$

Solving the quadratic equation $n^2 - 3n - 2446 = 0$: $n = \frac{3 \pm \sqrt{9 - 4(1)(-2446)}}{2} = \frac{3 \pm \sqrt{9793}}{2} \approx \frac{3 + 98.96}{2} \approx 50.98$

Thus, $n \le 50$.

Combining both bounds, the only integer value possible for $n$ is: $n = 50$

Finding $k$:

Substitute $n = 50$ back into the equation for $2k$: $2k = \frac{50 \times 51}{2} - 1225$ $2k = 1275 - 1225$ $2k = 50 \implies k = 25$

Evaluating $k - 20$:

$k - 20 = 25 - 20 = 5$