Math Problem Statement

You have a box of billiard balls. You arrange all but five balls in the shape of an equilateral triangle. Then you put all of the ballls back in the box and start over. This time, you try to make an equilartel triangle where each side has one more ball than in the frist arrangement. But this time you are 11 ball short! How many ball were originally in the box?

Solution

This problem can be understood by using triangular numbers. Triangular numbers represent how many balls can be arranged in an equilateral triangle, where each side contains a certain number of balls.

The formula for the nn-th triangular number TnT_n is given by:

Tn=n(n+1)2T_n = \frac{n(n+1)}{2}

where nn is the number of balls along one side of the equilateral triangle.

Step 1: The first arrangement

Let’s assume in the first arrangement, the number of balls on each side of the triangle is nn. The total number of balls arranged in the triangle is:

Tn=n(n+1)2T_n = \frac{n(n+1)}{2}

But we are told that not all the balls were used, specifically, 5 balls were left aside. So, the total number of balls in the box can be written as:

Total number of balls=Tn+5\text{Total number of balls} = T_n + 5

Step 2: The second arrangement

In the second arrangement, you attempt to create a triangle where each side has one more ball than in the first arrangement. So the number of balls along each side of this triangle is n+1n+1. The number of balls needed for this arrangement is:

Tn+1=(n+1)(n+2)2T_{n+1} = \frac{(n+1)(n+2)}{2}

However, you are 11 balls short. This means that the total number of balls in the box is 11 fewer than the number needed for this new arrangement:

Total number of balls=Tn+111\text{Total number of balls} = T_{n+1} - 11

Step 3: Equating the two expressions for the total number of balls

From Step 1 and Step 2, we have two expressions for the total number of balls in the box:

  1. Tn+5T_n + 5
  2. Tn+111T_{n+1} - 11

Setting these two equal gives:

Tn+5=Tn+111T_n + 5 = T_{n+1} - 11

Simplifying:

Tn+1Tn=16T_{n+1} - T_n = 16

Step 4: Finding the difference between consecutive triangular numbers

The difference between consecutive triangular numbers is:

Tn+1Tn=(n+1)T_{n+1} - T_n = (n+1)

So, we have:

n+1=16n + 1 = 16

Thus, n=15n = 15.

Step 5: Calculating the total number of balls

Now that we know n=15n = 15, we can calculate the total number of balls in the box.

First, find TnT_n:

T15=15(15+1)2=15×162=120T_{15} = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120

So, the total number of balls in the box is:

T15+5=120+5=125T_{15} + 5 = 120 + 5 = 125

Thus, the total number of balls originally in the box is 125.

Would you like more details on any part of this solution?

Here are 5 related questions to consider:

  1. What are triangular numbers, and how are they related to combinatorics?
  2. How can we generalize this problem to any other pattern of arrangement?
  3. What other shapes can we use for similar counting problems, and what are their formulas?
  4. How do differences between consecutive triangular numbers relate to arithmetic sequences?
  5. Can we solve this problem using algebraic manipulation instead of triangular numbers?

Tip: Triangular numbers are part of a larger family called "polygonal numbers," which can describe arrangements of objects in many geometric shapes.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Triangular Numbers
Arithmetic Sequences

Formulas

T_n = (n(n+1))/2 (Formula for the n-th triangular number)
T_{n+1} - T_n = (n+1) (Difference between consecutive triangular numbers)

Theorems

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Suitable Grade Level

Grades 8-10