Geometry Puzzle – Critical Thinking 19

Challenge:

N and P are prime numbers. How many divisors N\(^2\) × P\(^4\) has?

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If p and q are prime numbers, then for p\(^m\)q\(^n\), the number of divisors is (m+1)(n+1).
Thus the number of divisors for N\(^2\) × P\(^4\) is (2+1)(4+1) = 15
Another way to solve this problem is by providing numbers for N and P. Let’s put 2 for N and 3 for P. Then, the value of N\(^2\) × P\(^4\) is 2\(^2\) × 3\(^4\) = 324
Now, find the factors of 324.
324: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, 324
324 have 15 factors.

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Frequently Asked Questions

Why does the divisor formula multiply (exponent + 1) terms?

Each divisor of \( p_1^{a_1} \cdots p_k^{a_k} \) is uniquely determined by picking an exponent for each prime: \( 0, 1, 2, \ldots, a_i \) for \( p_i \). That is \( a_i + 1 \) choices for each prime. Multiplying gives the total.

What if N equals P in the puzzle?

Then \( N^2 \cdot P^4 = P^2 \cdot P^4 = P^6 \), a single prime raised to the sixth power. The divisor count becomes \( 6 + 1 = 7 \) — half the previous answer. The puzzle assumes distinct primes for the clean 15-divisor answer.

Can I see the 15 divisors written out for a concrete example?

Take \( N = 2 \), \( P = 3 \): \( 2^2 \cdot 3^4 = 4 \cdot 81 = 324 \). Its divisors are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, 324 — that is exactly 15.

What is prime factorization?

Writing a positive integer as a product of prime numbers raised to powers. Every integer greater than 1 has exactly one prime factorization (Fundamental Theorem of Arithmetic). For example, \( 72 = 2^3 \cdot 3^2 \).

Why must N and P be prime for the formula to give a single answer?

Composite numbers themselves split into smaller primes. If N or P were composite, the factorization of \( N^2 \cdot P^4 \) would involve more than two primes, and the divisor count would change. Primes lock down the count of distinct prime factors.

How is divisor counting used outside puzzles?

In number theory: studying perfect numbers, abundant/deficient numbers, and the divisor function are core topics. In cryptography: RSA security depends on the difficulty of factoring large integers, and divisor structure plays a role in choosing prime parameters.

How many divisors does N^2 * P^3 * Q^2 have, with three distinct primes?

Apply the same formula: \( (2+1)(3+1)(2+1) = 3 \cdot 4 \cdot 3 = 36 \) divisors. The formula extends to any number of distinct prime factors.

What is a divisor versus a factor?

They are the same thing in this context — a positive integer that divides the original number with no remainder. Both 1 and the number itself count as divisors.

How do I find the prime factorization of a number quickly?

Trial division by the smallest primes (2, 3, 5, 7, 11, …) until you reach 1. For 360: \( 360 = 2 \cdot 180 = 2^2 \cdot 90 = 2^3 \cdot 45 = 2^3 \cdot 3^2 \cdot 5 \). Stop checking primes once \( p^2 \) exceeds what is left.

Where does this puzzle fit in middle school math?

It sits naturally with units on factors, multiples, prime factorization, GCF/LCM, and exponent rules — typically Grades 6 through 8 in Common Core. Strong upper-elementary students can grasp the divisor count via examples.

Related Lessons You May Like

• How to find prime and composite numbers
• How to find the greatest common factor
• How to apply divisibility rules
• How to solve permutations and combinations
• How to solve multi-step word problems

If your student enjoys puzzles like this, Geometry for Beginners works the same relationships inside a full curriculum. For the number-theory side, Pre-Algebra for Beginners covers primes, factors, and divisibility from the ground up.