A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pa

Q: Wait, why do we reject the negative root when estimating $n$?

That is a fantastic question! In this problem, $n$ represents the total number of cards in a pack. By definition, a pack cannot have a negative number of cards. Therefore, we only consider the positive root of our quadratic estimation. Always remember to check the physical constraints of your variables!

Q: How do I know $n=50$ is the only possible value for $n$?

Great intuition. We found $n=50$ by testing values near our estimate. If we had chosen $n=49$, the total sum would be $\frac{49 \times 50}{2} = 1225$. Substituting this into our equation $\frac{n(n+1)}{2} - 2k = 1225$, we would get $1225 - 2k = 1225$, which implies $2k = 0$, or $k=0$. Since our cards are numbered from $1$ to $n$, $k$ cannot be $0$. Thus, $n=50$ is the unique solution that satisfies the condition $k \ge 1$.

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm! However, this is a pure number theory and sequence problem. While physics often uses conservation laws, here we are using the 'conservation' of the sum of the sequence. The total sum is fixed, and we are simply partitioning it. Stick to the algebraic properties of sequences, and you will find the path forward!

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 $k - 20 =$ .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding a deck of cards, numbered sequentially from $1$ to $n$. There is a beautiful, underlying order to this sequence, a rhythm defined by the sum of the first $n$ natural numbers.

We know this sum is given by the elegant formula:

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

This is our starting point, our anchor in the sea of numbers. Now, consider the twist: two consecutive cards are removed. If we call the smaller of these two cards $k$, then the next card is naturally $k+1$.

Their combined value is $k + (k+1) = 2k + 1$. This is the 'missing' piece of our sum.

### The Equation of the Gap

We are told that after removing these two cards, the sum of the remaining cards is $1224$. This creates a powerful bridge between our variables $n$ and $k$.

We can write this as:

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

By distributing the negative sign and moving the constant, we arrive at:

$$\frac{n(n+1)}{2} - 2k = 1225$$

This equation is the heart of the problem. It tells us that the total sum of the pack, $\frac{n(n+1)}{2}$, must be exactly $1225$ plus the value of $2k$.

Since $k$ is a card number, it must be at least $1$, meaning $2k$ is at least $2$. Therefore, the total sum $\frac{n(n+1)}{2}$ must be strictly greater than $1225$.

### The Art of Estimation

How do we find $n$ when we have two unknowns? We use the power of estimation. We know $\frac{n(n+1)}{2}$ is roughly $\frac{n^2}{2}$.

Setting $\frac{n^2}{2} \approx 1225$, we find $n^2 \approx 2450$. We are looking for a perfect square near $2450$.

We know $49^2 = 2401$ and $50^2 = 2500$. This suggests that $n$ is likely $50$.

Let us test $n=50$. The total sum is:

$$\frac{50 	imes 51}{2} = 1275$$

This is indeed greater than $1225$, which is a perfect sign.

### The Final Revelation

With $n=50$, our equation becomes $1275 - 2k = 1225$. Rearranging this, we get:

$$2k = 1275 - 1225 = 50$$

This leads us to $k = 25$. The removed cards were $25$ and $26$.

Everything fits! The cards are within the range $[1, 50]$, and the logic holds.

Finally, the question asks for $k - 20$. Substituting our value, we get $25 - 20 = 5$.

The final answer is **5**.

Enter Numerical Value:

Visualized Solution

Total Sum of $n$ Cards

Identifying the Removed Cards

Setting up the Equation

Simplifying the Equation

Estimating the Value of $n$

Testing Values for $n$

Solving for $k$

Verifying the Solution

Final Calculation: $k - 20$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Equation of the Gap

The Art of Estimation

The Final Revelation

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Enter Numerical Value:

Visualized Solution

Total Sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n Cards

Identifying the Removed Cards

Setting up the Equation

Simplifying the Equation

Estimating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Testing Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Verifying the Solution

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k−20