Question

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?

• A. 32
• B. 34
• C. 38
• D. 36
• E. 24

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.

Answer 1

The Correct Option is D

Step 1: Identify the Pattern:
Number of balls in layers: 1st layer: 1, 2nd: 3, 3rd: 6, 4th: 10, ... These are triangular numbers. The $n$th triangular number is $T_n = \frac{n(n+1)}{2}$. Check: $T_1=1$, $T_2=3$, $T_3=6$, $T_4=10$, etc. So, the number of balls in the $n$th layer is $\frac{n(n+1)}{2}$.
Step 2: Total Number of Balls:
Total balls = sum of triangular numbers from $n=1$ to $N$, where $N$ is the number of layers. $\sum_{n=1}^{N} \frac{n(n+1)}{2} = \frac{1}{2} \sum_{n=1}^{N} (n^2 + n) = \frac{1}{2} \left( \sum n^2 + \sum n \right)$. $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$. $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$. So total = $\frac{1}{2} \left( \frac{N(N+1)(2N+1)}{6} + \frac{N(N+1)}{2} \right)$. $= \frac{1}{2} \cdot N(N+1) \left( \frac{2N+1}{6} + \frac{1}{2} \right)$. $= \frac{N(N+1)}{2} \cdot \left( \frac{2N+1+3}{6} \right)$. $= \frac{N(N+1)}{2} \cdot \frac{2N+4}{6} = \frac{N(N+1)(2N+4)}{12}$. $= \frac{N(N+1)(N+2)}{6}$.
Step 3: Set Total Equal to 8436:
$\frac{N(N+1)(N+2)}{6} = 8436$. $N(N+1)(N+2) = 8436 \times 6 = 50616$.
Step 4: Approximate $N$:
$N^3 \approx 50616 \implies N \approx \sqrt[3]{50616} \approx 36.9$. Try $N=36$: $36 \times 37 \times 38 = 36 \times 1406 = 50616$. Exactly matches. $N=36$ works.
Step 5: Final Answer:
The number of horizontal layers is 36.