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Difficulty: HardPrime Factorization, GCD, and LCM

A positive integer nn has the prime factorization 2a⋅3b⋅5c2^a \cdot 3^b \cdot 5^c, where aa, bb, and cc are positive integers. Given that gcd⁡(n,7200)=360\gcd(n, 7{}200) = 360 and lcm(n,1080)=3240\text{lcm}(n, 1{}080) = 3{}240, what is the value of a+b+ca + b + c?

  1. A
    6
  2. B
    7
  3. 8Answer
  4. D
    9
  5. E
    10

Answer

8
By prime factorizing each given term, 7200=25⋅32⋅527{}200 = 2^5 \cdot 3^2 \cdot 5^2, 360=23⋅32⋅51360 = 2^3 \cdot 3^2 \cdot 5^1, 1080=23⋅33⋅511{}080 = 2^3 \cdot 3^3 \cdot 5^1, and 3240=23⋅34⋅513{}240 = 2^3 \cdot 3^4 \cdot 5^1. Applying the definition of GCD as taking the minimum exponent gives min⁡(a,5)=3  ⟹  a=3\min(a, 5) = 3 \implies a = 3 and min⁡(c,2)=1  ⟹  c=1\min(c, 2) = 1 \implies c = 1. Applying the definition of LCM as taking the maximum exponent gives max⁡(b,3)=4  ⟹  b=4\max(b, 3) = 4 \implies b = 4. Thus, a+b+c=3+4+1=8a + b + c = 3 + 4 + 1 = 8.

Step-by-Step Solution

1
Find the prime factorizations of the given integers and the GCD/LCM values.
7200=25⋅32⋅527{}200 = 2^5 \cdot 3^2 \cdot 5^2, 360=23⋅32⋅51360 = 2^3 \cdot 3^2 \cdot 5^1, 1080=23⋅33⋅511{}080 = 2^3 \cdot 3^3 \cdot 5^1, and 3240=23⋅34⋅513{}240 = 2^3 \cdot 3^4 \cdot 5^1.
Decomposing into prime factors allows direct comparison of exponents using GCD (minimum exponent) and LCM (maximum exponent) rules.
2
Analyze the GCD condition gcd⁡(n,7200)=360\gcd(n, 7{}200) = 360.
min⁡(a,5)=3  ⟹  a=3\min(a, 5) = 3 \implies a = 3, min⁡(b,2)=2  ⟹  b≥2\min(b, 2) = 2 \implies b \ge 2, and min⁡(c,2)=1  ⟹  c=1\min(c, 2) = 1 \implies c = 1.
The exponent of each prime factor in gcd⁡(x,y)\gcd(x, y) is the minimum of their respective exponents in xx and yy.
3
Analyze the LCM condition lcm(n,1080)=3240\text{lcm}(n, 1{}080) = 3{}240.
max⁡(a,3)=3  ⟹  a≤3\max(a, 3) = 3 \implies a \le 3, max⁡(b,3)=4  ⟹  b=4\max(b, 3) = 4 \implies b = 4, and max⁡(c,1)=1  ⟹  c≤1\max(c, 1) = 1 \implies c \le 1.
The exponent of each prime factor in lcm(x,y)\text{lcm}(x, y) is the maximum of their respective exponents in xx and yy.
4
Combine the exponent constraints to determine aa, bb, and cc, then calculate their sum.
a=3a = 3, b=4b = 4, and c=1c = 1, so a+b+c=3+4+1=8a + b + c = 3 + 4 + 1 = 8.
Combining a=3a=3, b≥2b \ge 2 with b=4b=4, and c=1c=1 with c≤1c \le 1 uniquely specifies (a,b,c)=(3,4,1)(a,b,c) = (3,4,1).

Key Concept

Prime Exponent Analysis of Greatest Common Divisor and Least Common Multiple
Estimated Time:2m 0s
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Topics

Prime Factorization, GCD, and LCM

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