02 // Visual Math & Finance SolversN = p₁^a₁ × p₂^a₂ × … × pₖ^aₖ, Divisors τ(N) = ∏(aᵢ + 1)
Visual Prime Factorization Tree & Divisor Lattice
By the Fundamental Theorem of Arithmetic, every integer $N > 1$ factors uniquely (up to ordering) into a product of prime powers $N = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, yielding $\tau(N) = \prod (a_i + 1)$ total positive divisors.
N = p₁^a₁ × p₂^a₂ × … × pₖ^aₖ, Divisors τ(N) = ∏(aᵢ + 1)Inputs & Parameters (2)
2±1999
2±1999
Prime Factorization (360)
2^3 × 3^2 × 5
6 prime leaves in tree
Total Divisors τ(360)
24 divisors
Smallest prime factor = 2
GCD(360, 150) [HCF]
30
Shared factors: 2×3×5
LCM(360, 150)
1,800
Check: 30 × 1,800 = 54,000
20-Row Comparison Table
Compare across:
| # | Target Integer (N) | Prime Factor Count Ω(N) | LCM(N, M) | Prime Factors of N |
|---|---|---|---|---|
| 01 | 2 | 2 | 150 | 2 |
| 02 | 54 | 6 | 1,350 | 2 × 3 × 3 × 3 |
| 03 | 107 | 1 | 16,050 | 107 |
| 04 | 159 | 3 | 7,950 | 3 × 53 |
| 05 | 212 | 2 | 15,900 | 2 × 2 × 53 |
| 06 | 264 | 6 | 6,600 | 2 × 2 × 2 × 3 × 11 |
| 07 | 317 | 1 | 47,550 | 317 |
| 08 | 369 | 3 | 18,450 | 3 × 3 × 41 |
| 09 | 422 | 2 | 31,650 | 2 × 211 |
| 10 | 474 | 6 | 11,850 | 2 × 3 × 79 |
| 11 | 527 | 1 | 79,050 | 17 × 31 |
| 12 | 579 | 3 | 28,950 | 3 × 193 |
| 13 | 632 | 2 | 47,400 | 2 × 2 × 2 × 79 |
| 14 | 684 | 6 | 17,100 | 2 × 2 × 3 × 3 × 19 |
| 15 | 737 | 1 | 110,550 | 11 × 67 |
| 16 | 789 | 3 | 39,450 | 3 × 263 |
| 17 | 842 | 2 | 63,150 | 2 × 421 |
| 18 | 894 | 6 | 22,350 | 2 × 3 × 149 |
| 19 | 947 | 1 | 142,050 | 947 |
| 20 | 999 | 3 | 49,950 | 3 × 3 × 3 × 37 |
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Key Analytical Takeaways
- 01Composite internal nodes (Cobalt outline) recursively split into balanced factor pairs until every leaf node is an indivisible prime number (Mineral Teal badge).
- 02$N = 360 = 2^3 \times 3^2 \times 5^1$ is a highly composite number possessing $(3+1)(2+1)(1+1) = 24$ distinct positive divisors—which is why circles have $360^\circ$.
- 03Regardless of which factor pair you split first, the multiset of prime leaf nodes at the bottom of the tree is invariant.
Frequently Asked Questions
Why is the number 1 not considered a prime number?+
If 1 were prime, prime factorization would no longer be unique because you could multiply by 1^k arbitrarily many times (e.g., 6 = 2×3 = 1×2×3 = 1×1×2×3).
How do you count all divisors of a number from its prime factorization?+
Add 1 to each prime factor's exponent and multiply them together: for 360 = 2³ × 3² × 5¹, the total number of divisors is (3+1)(2+1)(1+1) = 24.
Does the order in which I split factors change the final prime leaves?+
Never. By the Fundamental Theorem of Arithmetic, every integer greater than 1 has a unique multiset of prime factors regardless of which composite pair you split first.