Question

Difficulty: Very hardPrime Factorization, GCD, and LCM

Two positive integers aa and bb have prime factorizations of the form a=2x×3y×5za = 2^x \times 3^y \times 5^z and b=2u×3v×5wb = 2^u \times 3^v \times 5^w, where x,y,z,u,v,wx, y, z, u, v, w are non-negative integers. The greatest common divisor (GCD) of aa and bb is 180180, and the least common multiple (LCM) of aa and bb is 54,00054,000. If aa has exactly 2424 positive divisors, what is the value of bb?

Answer: 18000

Answer

18000
Prime factorizing GCD(a,b)=180=22×32×51\text{GCD}(a,b) = 180 = 2^2 \times 3^2 \times 5^1 and LCM(a,b)=54,000=24×33×53\text{LCM}(a,b) = 54,000 = 2^4 \times 3^3 \times 5^3 reveals that the pair of exponents for bases 2,3,52, 3, 5 are {2,4}\{2, 4\}, {2,3}\{2, 3\}, and {1,3}\{1, 3\} respectively. Applying the total divisor formula (x+1)(y+1)(z+1)=24(x+1)(y+1)(z+1) = 24 for aa, x+1x+1 must divide 2424. Since x{2,4}x \in \{2, 4\}, xx must be 22 (since 55 does not divide 2424). This leaves (y+1)(z+1)=8(y+1)(z+1) = 8. With y{2,3}y \in \{2, 3\}, y+1y+1 must be 44 (since 33 does not divide 88), giving y=3y = 3 and z=1z = 1. Thus, a=22×33×51a = 2^2 \times 3^3 \times 5^1. The remaining exponent choices for bb are u=4u = 4, v=2v = 2, and w=3w = 3, giving b=24×32×53=18,000b = 2^4 \times 3^2 \times 5^3 = 18,000.

Step-by-Step Solution

1
Factorize GCD and LCM into prime bases.
GCD = 2^2 * 3^2 * 5^1; LCM = 2^4 * 3^3 * 5^3.
GCD represents the minimum prime exponents and LCM represents the maximum prime exponents of a and b.
2
Set up candidate exponent pairs for primes 2, 3, and 5.
{x, u} = {2, 4}, {y, v} = {2, 3}, and {z, w} = {1, 3}.
For any prime p, min(exp_p(a), exp_p(b)) = exp_p(GCD) and max(exp_p(a), exp_p(b)) = exp_p(LCM).
3
Solve for x, y, and z using the divisor count formula d(a) = (x+1)(y+1)(z+1) = 24.
x = 2, y = 3, z = 1.
If x = 4, then x+1 = 5, which is not a factor of 24. Thus x = 2. Then (y+1)(z+1) = 8. Since y in {2, 3}, y+1 = 4 implies y = 3 and z = 1.
4
Determine the exponents of b and calculate b.
u = 4, v = 2, w = 3, so b = 2^4 * 3^2 * 5^3 = 18,000.
Complementary values from the sets {x,u}, {y,v}, and {z,w} give the exponents for b.

Key Concept

Relationship between prime factor exponents in GCD, LCM, and divisor counting functions.
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