Questions: If we have 52 marbles and make the biggest possible equilateral triangle, how many marbles are left over?

Solution Steps

To solve this problem, we need to find the largest equilateral triangle that can be formed with the given marbles. An equilateral triangle with side length \( n \) has \( \frac{n(n+1)}{2} \) marbles. We need to find the largest \( n \) such that \( \frac{n(n+1)}{2} \leq 52 \). After finding this \( n \), we calculate the number of marbles used and subtract it from 52 to find the leftover marbles.

To find the largest \( n \) such that the number of marbles used in an equilateral triangle is less than or equal to 52, we use the formula for the number of marbles in an equilateral triangle:

\[ \text{Marbles} = \frac{n(n+1)}{2} \]

We need to solve the inequality:

\[ \frac{n(n+1)}{2} \leq 52 \]

Multiplying both sides by 2 gives:

\[ n(n+1) \leq 104 \]

We can test integer values for \( n \):

• For \( n = 10 \): \( 10 \times 11 = 110 \) (too large)
• For \( n = 9 \): \( 9 \times 10 = 90 \) (valid)

Thus, the largest \( n \) is \( 9 \).

Now, we calculate the number of marbles used for \( n = 9 \):

\[ \text{Used Marbles} = \frac{9(9+1)}{2} = \frac{9 \times 10}{2} = 45 \]

Finally, we find the number of leftover marbles:

\[ \text{Leftover} = 52 - 45 = 7 \]

Final Answer