Distributivity and Algebra | A

Question 1

IT'S PUZZLE TIME!

Coin Conjoin

Arrange 10 coins in a triangle as shown in the figure below on the left. The task is to turn the triangle upside down by moving one coin at a time. How many moves are needed? What is the minimum number of moves?

A triangle of 3 coins can be inverted (turned upside down) with a single move, and a triangle of 6 coins can be inverted by moving 2 coins.

The 10-coin triangle can be flipped with just 3 moves; did you figure out how? Find out the minimum possible moves needed to flip the next bigger triangle having 15 coins. Try the same for bigger triangular numbers.

Is there a simple way to calculate the minimum number of coin moves needed for any such triangular arrangement?

Solution

Understand the Question

A triangular arrangement of coins with $n$ rows contains a triangular number of coins given by: $N = \dfrac{n(n+1)}{2}$
Inverting (turning upside down) a triangular arrangement of $n$ rows can be achieved by moving the minimum number of coins, which follows the pattern: $\text{Minimum moves} = n - 1$
For any given number of coins $N$ , we can determine the number of rows $n$ and calculate the required moves.

(i) Find the minimum number of moves needed to turn the 10-coin triangle upside down.

Step 1 · Observe Pattern and Calculate Moves for 10 Coins

The total number of coins in a triangle with $n$ rows is $N = \dfrac{n(n+1)}{2}$ .

Comparing rows ( $n$ ) and minimum moves:

For $3$ coins: $n = 2 \implies \text{moves} = 1$
For $6$ coins: $n = 3 \implies \text{moves} = 2$
For $10$ coins: $n = 4 \implies \text{moves} = 3$
For $3$ coins ( $n = 2$ ):
- Move top coin $\text{C1}$ below $\text{C2}$ and $\text{C3}$ .
- Moves required $= 1$ .

Diagram 2

For $6$ coins ( $n = 3$ ):
- Move top coin $\text{C1}$ and bottom-right coin $\text{C6}$ .
- Moves required $= 2$ .

Diagram 3

For $10$ coins ( $n = 4$ ):
- Move top coin $\text{C1}$ , bottom-left coin $\text{C7}$ , and bottom-right coin $\text{C10}$ .
- Moves required $= 3$ .

$\text{Minimum moves for } 10 \text{ coins} = n - 1 = 4 - 1 = 3$

Answer

(i) $3$

(ii) Find the minimum possible moves needed to flip the next bigger triangle having 15 coins.

Step 1 · Find Number of Rows and Compute Moves for 15 Coins

Given total coins $N = 15$ .

\begin{aligned} \dfrac{n(n+1)}{2} &= 15 \\[0.6em] n(n+1) &= 30 \end{aligned}

Since $5 \times (5 + 1) = 5 \times 6 = 30$ , the number of rows is $n = 5$ .

\begin{aligned} \text{Minimum moves} &= n - 1 \\[0.6em] &= 5 - 1 \\[0.6em] &= 4 \end{aligned}

Answer

(ii) $4$

(iii) Find a simple way to calculate the minimum number of coin moves needed for any such triangular arrangement.

Step 1 · Derive General Formula in Terms of Total Coins

For a triangle with $n$ rows, the minimum moves required is $n - 1$ .

To express this in terms of total coins $N$ :

\begin{aligned} N &= \dfrac{n(n+1)}{2} \\[0.6em] 2N &= n^2 + n \\[0.6em] n^2 + n - 2N &= 0 \end{aligned}

Using the quadratic formula $n = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1, b = 1, c = -2N$ :

\begin{aligned} n &= \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-2N)}}{2(1)} \\[1.1em] &= \dfrac{-1 + \sqrt{1 + 8N}}{2} \quad (\text{since } n > 0) \end{aligned}

Substituting $n$ into $\text{Minimum moves} = n - 1$ :

\begin{aligned} \text{Minimum moves} &= \dfrac{-1 + \sqrt{1 + 8N}}{2} - 1 \\[1.1em] &= \dfrac{-1 + \sqrt{1 + 8N} - 2}{2} \\[1.1em] &= \dfrac{\sqrt{1 + 8N} - 3}{2} \end{aligned}

Answer

(iii) $n - 1 \text{ (where } n \text{ is the number of rows), or } \dfrac{\sqrt{1 + 8N} - 3}{2} \text{ (where } N \text{ is the total number of coins)}$

Common Mistakes

Moving Interior Coins: Attempting to move interior coins instead of corner/boundary coins, which leads to extra moves.
Confusing $n$ with $N$ : Using the total number of coins $N$ in place of the number of rows $n$ in the simple formula $n - 1$ .