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Question

Difficulty: MediumDivisibility, Factors, and Multiples

If nn is a positive integer such that 140n140n is a perfect square and 105n105n is a perfect cube, what is the minimum possible number of positive divisors of nn?

  1. A
    12
  2. B
    27
  3. 108Answer
  4. D
    756
  5. E
    15

Answer

108
To minimize the number of divisors of nn, express n=2a⋅3b⋅5c⋅7dn = 2^a \cdot 3^b \cdot 5^c \cdot 7^d. The condition that 140n=2a+2⋅3b⋅5c+1⋅7d+1140n = 2^{a+2} \cdot 3^b \cdot 5^{c+1} \cdot 7^{d+1} is a perfect square implies aa and bb must be even, while cc and dd must be odd. The condition that 105n=2a⋅3b+1⋅5c+1⋅7d+1105n = 2^a \cdot 3^{b+1} \cdot 5^{c+1} \cdot 7^{d+1} is a perfect cube requires aa to be a multiple of 3, and b,c,db, c, d to leave a remainder of 2 when divided by 3. The minimum non-negative integers satisfying both requirements are a=0a=0 (since 0 is even and a multiple of 3), b=2b=2 (even and leaves remainder 2 mod 3), c=5c=5 (odd and leaves remainder 2 mod 3), and d=5d=5 (odd and leaves remainder 2 mod 3). The minimum number of positive divisors is (0+1)(2+1)(5+1)(5+1)=108(0+1)(2+1)(5+1)(5+1) = 108.

Step-by-Step Solution

1
Find the prime factorizations of 140 and 105
140=22⋅5⋅7140 = 2^2 \cdot 5 \cdot 7 and 105=3⋅5⋅7105 = 3 \cdot 5 \cdot 7
Decomposing into prime factors allows analyzing exponent constraints for perfect powers.
2
Express nn in terms of prime factors n=2a⋅3b⋅5c⋅7dn = 2^a \cdot 3^b \cdot 5^c \cdot 7^d and determine constraints for 140n140n to be a perfect square
140n=2a+2⋅3b⋅5c+1⋅7d+1140n = 2^{a+2} \cdot 3^b \cdot 5^{c+1} \cdot 7^{d+1} requires aa to be even, bb to be even, cc to be odd, and dd to be odd.
All prime factor exponents in a perfect square must be even numbers.
3
Determine constraints for 105n105n to be a perfect cube
105n=2a⋅3b+1⋅5c+1⋅7d+1105n = 2^a \cdot 3^{b+1} \cdot 5^{c+1} \cdot 7^{d+1} requires a≡0(mod3)a \equiv 0 \pmod 3, b≡2(mod3)b \equiv 2 \pmod 3, c≡2(mod3)c \equiv 2 \pmod 3, and d≡2(mod3)d \equiv 2 \pmod 3.
All prime factor exponents in a perfect cube must be multiples of 3.
4
Find the smallest non-negative integers satisfying both sets of constraints for each exponent
a=0a = 0, b=2b = 2, c=5c = 5, and d=5d = 5
For aa: smallest non-negative even multiple of 3 is 0. For bb: smallest non-negative even number congruent to 2(mod3)2 \pmod 3 is 2. For cc and dd: smallest odd numbers congruent to 2(mod3)2 \pmod 3 are 5.
5
Calculate the total number of positive divisors of n=32⋅55⋅75n = 3^2 \cdot 5^5 \cdot 7^5
(2+1)(5+1)(5+1)=3×6×6=108(2 + 1)(5 + 1)(5 + 1) = 3 \times 6 \times 6 = 108
The divisor counting formula multiplies (ei+1)(e_i + 1) for each prime exponent eie_i.

Key Concept

Prime factor exponent constraints for perfect powers and the divisor counting formula
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Topics

Divisibility, Factors, and Multiples

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