There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile in this case will be how many?
Show Hint
The sum of the first $N$ triangular numbers is $\frac{N(N+1)(N+2)}{6}$. This is also the $N$th tetrahedral number.
Step 1: Understanding the Concept:
The balls are in triangular layers. The sum of the first N triangular numbers is a Tetrahedral number. Step 2: Key Formula or Approach:
Sum of first N triangular numbers \( = \frac{N(N+1)(N+2)}{6} \). Step 3: Detailed Explanation:
We need \( \frac{N(N+1)(N+2)}{6} = 8436 \).
\( N(N+1)(N+2) = 50616 \).
Find a cube root approximation: \( N^3 \approx 50616 \).
\( 30^3 = 27000, 40^3 = 64000 \). The answer is near 36-38.
Testing \( N = 36 \):
\( 36 \cdot 37 \cdot 38 = 1332 \cdot 38 = 50616 \). Matches perfectly. Step 4: Final Answer:
The number of layers is 36.