Question

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=

Answer 1

5

Answer 2

Let $n$ be the total number of cards in the pack, numbered from 1 to $n$. The sum of the numbers on these cards is $S_n = \frac{n(n+1)}{2}$.

Two consecutive numbered cards are removed from the pack. Let the numbers on these cards be $k$ and $k+1$. Since these cards are from the pack, we must have $1 \le k$ and $k+1 \le n$. This implies $1 \le k \le n-1$.

The sum of the numbers on the remaining cards is given as 1224. The sum of the remaining cards is equal to the total sum minus the sum of the removed cards. Sum of remaining cards = $S_n - (k + (k+1))$ $1224 = \frac{n(n+1)}{2} - (2k+1)$ $1224 = \frac{n(n+1)}{2} - 2k - 1$ $1225 = \frac{n(n+1)}{2} - 2k$ Multiply by 2: $2450 = n(n+1) - 4k$ $n(n+1) = 2450 + 4k$

We know that $1 \le k \le n-1$. We can use these bounds for $k$ to find the possible range for $n$. Since $k \ge 1$, we have $4k \ge 4$. $n(n+1) = 2450 + 4k \ge 2450 + 4(1) = 2454$. So, $n(n+1) \ge 2454$. Let's test values of $n$. For $n=49$, $n(n+1) = 49 \times 50 = 2450$, which is less than 2454. For $n=50$, $n(n+1) = 50 \times 51 = 2550$, which is greater than 2454. So, $n$ must be an integer $n \ge 50$.

Since $k \le n-1$, we have $4k \le 4(n-1) = 4n - 4$. $n(n+1) = 2450 + 4k \le 2450 + 4(n-1) = 2450 + 4n - 4 = 2446 + 4n$. So, $n(n+1) \le 2446 + 4n$. $n^2 + n \le 2446 + 4n$ $n^2 - 3n - 2446 \le 0$. To find the values of $n$ that satisfy this inequality, we find the roots of the quadratic equation $n^2 - 3n - 2446 = 0$. Using the quadratic formula, $n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-2446)}}{2(1)} = \frac{3 \pm \sqrt{9 + 9784}}{2} = \frac{3 \pm \sqrt{9793}}{2}$. The value of $\sqrt{9793}$ is approximately $98.96$. The roots are approximately $n_1 = \frac{3 - 98.96}{2} \approx -47.98$ and $n_2 = \frac{3 + 98.96}{2} \approx 50.98$. The inequality $n^2 - 3n - 2446 \le 0$ holds for values of $n$ between the roots, i.e., approximately $-47.98 \le n \le 50.98$. Since $n$ must be a positive integer, $n$ must satisfy $1 \le n \le 50$.

Combining the two conditions for $n$: $n \ge 50$ and $n \le 50.98$. The only integer value of $n$ that satisfies both conditions is $n=50$.

Now that we have $n=50$, we can find the value of $k$. Substitute $n=50$ into the equation $n(n+1) = 2450 + 4k$: $50(50+1) = 2450 + 4k$ $50 \times 51 = 2450 + 4k$ $2550 = 2450 + 4k$ $4k = 2550 - 2450$ $4k = 100$ $k = 25$.

We must check if this value of $k$ is valid for $n=50$. The condition is $1 \le k \le n-1$. $1 \le 25 \le 50-1 = 49$. The value $k=25$ is within the valid range.

The smaller of the numbers on the removed cards is $k$. We are asked to find the value of $k-20$. $k-20 = 25 - 20 = 5$.