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Difficulty: MediumDivisibility, Factors, and Multiples

A total of 360 identical notebooks are to be distributed equally among nn students, where n>1n > 1, such that each student receives an integer number of notebooks strictly greater than 1. If nn must be a multiple of 4, how many different values of nn are possible?

Answer: 11

Answer

The total number of possible values for nn is 11.
To find the number of valid values of nn, prime factorize 360=23×32×51360 = 2^3 \times 3^2 \times 5^1. For nn to be a factor of 360 and a multiple of 4, the exponent of 2 must be 2 or 3 (2 choices), the exponent of 3 can be 0, 1, or 2 (3 choices), and the exponent of 5 can be 0 or 1 (2 choices), yielding 2×3×2=122 \times 3 \times 2 = 12 total factors of 360 that are multiples of 4. Excluding n=360n = 360, which results in 1 notebook per student, leaves 12−1=1112 - 1 = 11 valid values.

Step-by-Step Solution

1
Prime factorize the total number of notebooks, 360.
360=23×32×51360 = 2^3 \times 3^2 \times 5^1
Prime factorization allows us to systematically count factors meeting specific divisibility conditions.
2
Calculate the number of factors of 360 that are multiples of 4.
2 choices for the exponent of 2 (222^2 or 232^3), 3 choices for the exponent of 3 (30,31,323^0, 3^1, 3^2), and 2 choices for the exponent of 5 (50,515^0, 5^1). Total factors = 2×3×2=122 \times 3 \times 2 = 12.
A factor is a multiple of 4 if and only if its prime factorization contains at least two factors of 2.
3
Apply the constraint that each student receives strictly more than 1 notebook.
If n=360n = 360, each student receives 360360=1\frac{360}{360} = 1 notebook, which violates the condition. Subtracting this case yields 12−1=1112 - 1 = 11 valid values.
The problem requires each student to receive an integer number of notebooks strictly greater than 1.

Key Concept

Counting constrained factors using prime factorization
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Topics

Divisibility, Factors, and Multiples

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