Question

Difficulty: Very hardFactors, Multiples, and Prime Factorization

Let SS be the set of all positive integers nn such that the least common multiple of nn and 4040 is 360360, and the greatest common divisor of nn and 100100 is a prime number. What is the sum of all possible values of nn in set SS?

  1. A
    18
  2. B
    45
  3. 63Answer
  4. D
    72
  5. E
    153

Answer

63
To find the sum of all possible values of nn, we first analyze the prime factorizations: 40=23×540 = 2^3 \times 5 and 360=23×32×5360 = 2^3 \times 3^2 \times 5. The least common multiple of nn and 4040 is 360360, meaning nn must be of the form 2a×32×5c2^a \times 3^2 \times 5^c where 0a30 \le a \le 3 and 0c10 \le c \le 1. Testing the candidate values against the condition that GCD(n,100)\text{GCD}(n, 100) must be a prime number (where 100=22×52100 = 2^2 \times 5^2) reveals that only n=18n = 18 (with GCF of 22) and n=45n = 45 (with GCF of 55) satisfy the conditions. The sum of these values is 18+45=6318 + 45 = 63.

Step-by-Step Solution

1
Find the prime factorizations of the given numbers: 4040 and 360360.
40=23×540 = 2^3 \times 5 and 360=23×32×5360 = 2^3 \times 3^2 \times 5.
Expressing the numbers in their prime factorizations helps determine the required prime factors of nn to satisfy the least common multiple condition.
2
Determine the constraints on the prime factorization of nn based on LCM(n,40)=360\text{LCM}(n, 40) = 360.
nn must be of the form 2a×32×5c2^a \times 3^2 \times 5^c, where 0a30 \le a \le 3 and 0c10 \le c \le 1.
Since the least common multiple is the product of the highest powers of all prime factors present in either number, nn must provide 323^2 (as 4040 has 303^0), can have 2a2^a up to 232^3, and can have 5c5^c up to 515^1.
3
List all 8 possible candidate values for nn.
For c=0c=0: n{9,18,36,72}n \in \{9, 18, 36, 72\}. For c=1c=1: n{45,90,180,360}n \in \{45, 90, 180, 360\}.
Evaluating all combinations of the exponents aa and cc generates the complete list of candidates.
4
Calculate the greatest common divisor of each candidate nn with 100=22×52100 = 2^2 \times 5^2, and check if the result is prime.
GCD(9,100)=1\text{GCD}(9, 100) = 1 (not prime); GCD(18,100)=2\text{GCD}(18, 100) = 2 (prime); GCD(36,100)=4\text{GCD}(36, 100) = 4 (not prime); GCD(72,100)=4\text{GCD}(72, 100) = 4 (not prime); GCD(45,100)=5\text{GCD}(45, 100) = 5 (prime); GCD(90,100)=10\text{GCD}(90, 100) = 10 (not prime); GCD(180,100)=20\text{GCD}(180, 100) = 20 (not prime); GCD(360,100)=20\text{GCD}(360, 100) = 20 (not prime). The valid values for nn are 1818 and 4545.
Only n=18n = 18 and n=45n = 45 satisfy the condition that their greatest common divisor with 100100 is a prime number (22 and 55, respectively).
5
Calculate the sum of the possible values of nn.
18+45=6318 + 45 = 63.
Summing the identified valid values yields the final answer.

Key Concept

Determining possible values of an unknown integer using prime factorization constraints from least common multiple (LCM) and greatest common divisor (GCD) conditions.
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