Understanding the Problem

A pack has $n$ cards numbered from 1 to $n$. Two consecutive cards are removed. Let the smaller number on the removed cards be $k$, so the removed numbers are $k$ and $k+1$. The sum of the remaining $n-2$ cards is 1224. We need to find $k-20$.

Step 1: Total Sum of All Cards

The sum of numbers from 1 to $n$ is given by the formula:

$S=\frac{n(n+1)}{2}$

Step 2: Sum After Removing Two Cards

After removing $k$ and $k+1$, the sum of the remaining cards is:

$\frac{n(n+1)}{2}-(k+k+1)=1224$

Simplify the removal:

$\frac{n(n+1)}{2}-(2k+1)=1224$

Step 3: Rearranging the Equation

Multiply both sides by 2 to eliminate the denominator:

$n(n+1)-2(2k+1)=2448$

Which simplifies to:

$n(n+1)-4k-2=2448$

Bring all terms to one side:

$n(n+1)-4k=2450$

Therefore:

$4k=n(n+1)-2450$

Since $k$ must be a positive integer, $n(n+1)$ must be greater than 2450 and the right side must be divisible by 4.

Step 4: Estimating n

We know $n(n+1)\approx n^2$. Solve $n^2\approx 2450$.

$\sqrt{2450}\approx 49.5$, so $n$ is around 49 or 50.

Let's test $n=50$:

$n(n+1)=50\times 51=2550$

Plug into the equation:

$4k=2550-2450=100$

Therefore, $k=\frac{100}{4}=25$

This gives a valid integer value for $k$.

Step 5: Verification

Total sum for n=50: $\frac{50\times 51}{2}=1275$

Sum removed: $k+(k+1)=25+26=51$

Remaining sum: $1275-51=1224$, which matches the given condition.

Final Answer

We are asked to find $k-20$.

Since $k=25$, then:

$k-20=25-20=5$

Related Topics & Formulae

Sum of an Arithmetic Series: The sum of the first $n$ natural numbers is given by $S=\frac{n(n+1)}{2}$.

Problem-Solving Strategy: For problems involving the removal of items from a set, a common approach is to calculate the total sum and subtract the sum of the removed items to equate it to the given remaining sum. The key is to ensure the values for $n$ and $k$ are positive integers that satisfy all conditions.