A-1

Chapter: DISTRIBUTIVITY AND ALGEBRA
Class: 8 (Class 8)
Category: activity

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Question

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?

Question diagram(s):

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The minimum number of moves to invert a triangle of coins is related to the number of rows it has.

Step 1 — Understanding the Pattern

Let us represent the number of rows in a triangle of coins by $n$. The total number of coins in such a triangle is a triangular number, $N = \frac{n(n+1)}{2}$.

We are given some examples:

• For 3 coins, $n=2$ (since $2(2+1)/2 = 3$). The problem states 1 move is needed.
• For 6 coins, $n=3$ (since $3(3+1)/2 = 6$). The problem states 2 moves are needed.
• For 10 coins, $n=4$ (since $4(4+1)/2 = 10$). The problem states 3 moves are needed.

Let us observe the relationship between the number of rows ($n$) and the minimum number of moves.

• If $n=2$, moves = 1.
• If $n=3$, moves = 2.
• If $n=4$, moves = 3.

We can see a clear pattern here: the minimum number of moves is always one less than the number of rows. So, the minimum number of moves = $n-1$.

Let's visualize how these moves work for each case:

• For 3 coins (n=2):
  • Original:
    • C1
    • C2 C3
  • Move C1 (the top coin) to the position below C2 and C3.
  • The coins C2 and C3 stay in their places, forming the new top row.
  • Resulting inverted triangle:
    • C2 C3
    • C1
  • This requires 1 move.

• For 6 coins (n=3):
  • Original:
    • C1
    • C2 C3
    • C4 C5 C6
  • Move C1 (the top coin) to the position below C4 and C5.
  • Move C6 (the bottom-right coin) to the position above C2 and C4.
  • The coins C2, C3, C4, C5 stay in their places.
  • Resulting inverted triangle:
    • C6
    • C2 C3
    • C4 C5 C1
  • This requires 2 moves.

• For 10 coins (n=4):
  • Original:
    • C1
    • C2 C3
    • C4 C5 C6
    • C7 C8 C9 C10
  • Move C1 (the top coin) to the position below C8 and C9.
  • Move C7 (the bottom-left coin) to the position above C2 and C4.
  • Move C10 (the bottom-right coin) to the position above C3 and C6.
  • The coins C2, C3, C4, C5, C6, C8, C9 stay in their places.
  • Resulting inverted triangle:
    • C10 C4 C2 C7
    • . C6 C5 C3
    • . . C9 C8
    • . . . C1
  • This requires 3 moves.

The minimum number of moves for a triangle with $n$ rows is indeed $n-1$.

$\boxed{\text{Minimum moves for 10 coins} = \mathbf{3}}$

Step 2 — Minimum moves for 15 coins

First, we need to find the number of rows ($n$) for a triangle with 15 coins. Let $N$ be the total number of coins, so $N = 15$. The formula for the total number of coins is $N = \frac{n(n+1)}{2}$.

$15 = \frac{n(n+1)}{2}$

Let us multiply both sides by 2:

$30 = n(n+1)$

We need to find a number $n$ such that when multiplied by the next consecutive number ($n+1$), the result is 30. We can try values for $n$: If $n=4$, $4 \times (4+1) = 4 \times 5 = 20$. This is too small. If $n=5$, $5 \times (5+1) = 5 \times 6 = 30$. This is correct! So, for 15 coins, the number of rows $n$ is 5.

Now, we use the pattern we found in Step 1: the minimum number of moves is $n-1$.

$\text{Minimum moves} = n-1$

$\text{Minimum moves} = 5-1$

$= \mathbf{4}$

$\boxed{\text{Minimum moves for 15 coins} = \mathbf{4}}$

Step 3 — General formula for any triangular arrangement

Let $N$ be the total number of coins in the triangular arrangement. Let $n$ be the number of rows in this triangle. We know that the total number of coins $N$ is given by the formula for triangular numbers:

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

From Step 1, we found that the minimum number of moves required to invert the triangle is $n-1$. To find a general formula in terms of $N$, we first need to express $n$ in terms of $N$. Let us rearrange the formula for $N$:

$2N = n(n+1)$

$2N = n^2 + n$

Let us move $2N$ to the right side to form a quadratic equation:

$n^2 + n - 2N = 0$

We can solve for $n$ using the quadratic formula, which states that for an equation $ax^2 + bx + c = 0$, the solutions for $x$ are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $x=n$, $a=1$, $b=1$, and $c=-2N$.

$n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2N)}}{2(1)}$

$n = \frac{-1 \pm \sqrt{1 + 8N}}{2}$

Since $n$ represents the number of rows, it must be a positive value. So, we take the positive square root:

$n = \frac{-1 + \sqrt{1 + 8N}}{2}$

Now, we substitute this expression for $n$ into our formula for the minimum number of moves, which is $n-1$:

$\text{Minimum moves} = n-1$

$\text{Minimum moves} = \frac{-1 + \sqrt{1 + 8N}}{2} - 1$

To combine these terms, we find a common denominator:

$\text{Minimum moves} = \frac{-1 + \sqrt{1 + 8N}}{2} - \frac{2}{2}$

$\text{Minimum moves} = \frac{-1 + \sqrt{1 + 8N} - 2}{2}$

$\text{Minimum moves} = \frac{\sqrt{1 + 8N} - 3}{2}$

This formula gives the minimum number of moves needed for any triangular arrangement with $N$ coins.

$\boxed{\text{General formula for minimum moves} = \frac{\sqrt{1 + 8N} - 3}{2}}$

Answer

(i) The minimum number of moves needed to flip the 10-coin triangle is 3. (ii) The minimum number of moves needed to flip the next bigger triangle having 15 coins is 4. (iii) A simple way to calculate the minimum number of coin moves needed for any such triangular arrangement is $n-1$, where $n$ is the number of rows. This can also be expressed as $\frac{\sqrt{1 + 8N} - 3}{2}$, where $N$ is the total number of coins.