2023 AMC 12B Problem 24

Question

Suppose that

a,

b,

c,

and

d

are positive integers satisfying all of the following relations.

\begin{aligned} abcd &= 2^6\cdot 3^9\cdot 5^7\\ \operatorname{lcm}(a,b) &= 2^3\cdot 3^2\cdot 5^3\\ \operatorname{lcm}(a,c) &= 2^3\cdot 3^3\cdot 5^3\\ \operatorname{lcm}(a,d) &= 2^3\cdot 3^3\cdot 5^3\\ \operatorname{lcm}(b,c) &= 2^1\cdot 3^3\cdot 5^2\\ \operatorname{lcm}(b,d) &= 2^2\cdot 3^3\cdot 5^2\\ \operatorname{lcm}(c,d) &= 2^2\cdot 3^3\cdot 5^2 \end{aligned}

What is

\gcd(a,b,c,d)?

30

45

3

15

6

Answer: C

Concepts:least common multiple greatest common divisor prime factorization

Difficulty rating: 2270

Small hint:

Work one prime at a time;

\gcd

uses the minimum exponent and

\operatorname{lcm}

the maximum

Big hint:

For each prime, pin down the four exponents from the max (lcm) and total (product) conditions, then take the minimum

Solution:

Handle each prime separately using the exponents of

a,b,c,d.

Prime $2$ (total

6

):

\max(b,c)=1

forces

a=3;

then

b+c+d=3

with

\max(b,d)=\max(c,d)=2

gives

d=2

and

\{b,c\}=\{0,1\},

so the minimum exponent is

0.

Prime $3$ (total

9

):

\max(a,b)=2

with the other lcms equal to

3

forces

c=d=3;

then

a+b=3

with

\max(a,b)=2

gives

\{a,b\}=\{1,2\},

so the minimum is

1.

Prime $5$ (total

7

):

\max(a,b)=3

with

\max(b,c),

\max(b,d),\max(c,d)=2

forces

a=3;

then

b+c+d=4

with each

\le 2

and pairwise maxima

2

gives two of them equal to

2

and one equal to

0,

so the minimum is

0.

Therefore

\gcd(a,b,c,d)=2^0\cdot 3^1\cdot 5^0=3.

Thus, the correct answer is C.

2023 AMC 12B Problem 24

Answer: C

Difficulty rating: 2270

Small hint:

Big hint:

Solution:

Problem 24 in other years