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Question

Difficulty: MediumFactors, Multiples, and Prime Factorization

What is the total number of positive divisors of the integer 1080010800 that are perfect squares?

Answer: 12

Answer

12
The correct answer is found by prime factorizing 1080010800 into 24×33×522^4 \times 3^3 \times 5^2. A factor is a perfect square only when its prime exponents are all even integers. Counting the available even exponents from 0 up to the maximum exponent for each prime gives 3 choices for base 2 (0, 2, 4), 2 choices for base 3 (0, 2), and 2 choices for base 5 (0, 2). Multiplying these independent choices gives 3×2×2=123 \times 2 \times 2 = 12.

Step-by-Step Solution

1
Find the prime factorization of 1080010800.
10800=108×100=(22×33)×(22×52)=24×33×5210800 = 108 \times 100 = (2^2 \times 3^3) \times (2^2 \times 5^2) = 2^4 \times 3^3 \times 5^2
The prime factorization is necessary to systematically identify and count the divisors of the number based on exponent combinations.
2
Identify the mathematical condition for a divisor to be a perfect square.
A divisor D=2a×3b×5cD = 2^a \times 3^b \times 5^c is a perfect square if and only if the exponents aa, bb, and cc are all even integers.
Perfect squares must have even exponents in their prime factorization so that their square roots are integers.
3
List the valid even exponent values for each prime base.
For base 2 (max exponent 4), valid even values are 0, 2, 4 (3 options). For base 3 (max exponent 3), valid even values are 0, 2 (2 options). For base 5 (max exponent 2), valid even values are 0, 2 (2 options).
These are the exponents that keep the divisor as a perfect square while still remaining a factor of 1080010800.
4
Calculate the total number of perfect square divisors.
3×2×2=123 \times 2 \times 2 = 12
Applying the fundamental counting principle, we multiply the number of independent choices for each prime exponent.

Key Concept

Identifying perfect square factors using prime factorization
Estimated Time:1m 30s
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Topics

Factors, Multiples, and Prime Factorization

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