Question

The GCD of 6480, 7200 and a third number is 144. The LCM of the three numbers is 2^5×3^5×5^2×7^3. Find the smallest third number as a product of its prime factors

Answer 1

To find the smallest third number \( x \) given that the GCD of \( 6480 \), \( 7200 \), and \( x \) is \( 144 \), and that the LCM of these three numbers is \( 2^5 \times 3^5 \times 5^2 \times 7^3 \), we will use the following properties of GCD and LCM:

1. The relationship between GCD and LCM:
\[
\text{GCD}(a, b, c) \times \text{LCM}(a, b, c) = a \times b \times c
\]

Here, \( a = 6480 \), \( b = 7200 \), and \( c = x \).

First, let's calculate \( a = 6480 \) and \( b = 7200 \) in terms of their prime factorizations:

### Step 1: Prime Factorization of \( 6480 \)

\[
6480 = 648 \times 10 = (81 \times 8) \times (10) = (3^4 \times 2^3) \times (2 \times 5) = 2^4 \times 3^4 \times 5
\]

Calculating \( 6480 \):
1. \( 6480 \div 10 = 648 \).
2. \( 648 \div 9 = 72 \), so now we find factors of \( 72 \): \( 72 = 8 \times 9 = 2^3 \times 3^2 \).
3. Including the factors of \( 10 \) (i.e., \( 2 \times 5 \)):
\[
6480 = 2^4 \times 3^4 \times 5
\]

### Step 2: Prime Factorization of \( 7200 \)

\[
7200 = 720 \times 10 = (72 \times 10) \times (10) = (8 \times 9) \times (10) \times (10) = (2^3 \times 3^2) \times (2 \times 5) \times (2 \times 5)
\]

Calculating \( 7200 \):
1. \( 7200 \div 100 = 72 \), so \( 7200 = 72 \times 100 \) where \( 100 = 10^2 = 2^4 \times 5^2 \).
2. Including previously found factors:
\[
7200 = (2^3 \times 3^2) \times (2^4 \times 5^2) = 2^7 \times 3^2 \times 5^2
\]

### Step 3: Find GCD and LCM

From above, we have:
\[
6480 = 2^4 \times 3^4 \times 5^1
\]
\[
7200 = 2^7 \times 3^2 \times 5^2
\]

The GCD can be computed as:
\[
\text{GCD}(6480, 7200) = 2^{\min(4, 7)} \times 3^{\min(4, 2)} \times 5^{\min(1, 2)} = 2^4 \times 3^2 \times 5^1 = 240
\]

Next, we compute the GCD of \( 6480, 7200, x \):
\[
\text{GCD}(6480, 7200, x) = 144 = 2^4 \times 3^2
\]

### Step 4: Set Up the Third Number \( x \)

Let \( x \) be in the form:
\[
x = 2^a \times 3^b \times 5^c \times 7^d
\]

Given:
- To satisfy GCD, we have:
\[
\min(4, a) = 4 \quad \Rightarrow \quad a \geq 4
\]
\[
\min(4, b) = 2 \quad \Rightarrow \quad b \geq 2
\]
\[
c \text{ must be } \geq 0
\]
\[
d \text{ may be } 0 \text{ (since it isn't part of previous numbers)}
\]

### Step 5: Calculate LCM

Using LCM:
\[
\text{LCM}(6480, 7200, x) = 2^{\max(4, a)} \times 3^{\max(4, b)} \times 5^{\max(1, c)} \times 7^{\max(0, d)} = 2^5 \times 3^5 \times 5^2 \times 7^3
\]

From this, we derive:
- \( \max(4, a) = 5 \quad \Rightarrow a = 5 \)
- \( \max(4, b) = 5 \quad \Rightarrow b = 5 \)
- \( \max(1, c) = 2 \quad \Rightarrow c = 2 \)
- \( \max(0, d) = 3 \quad \Rightarrow d = 3 \)

Thus:
\[
x = 2^5 \times 3^5 \times 5^2 \times 7^3
\]

### Final Step: Smallest \( x \)

To ensure minimum values for \( a, b, c, d \):
\[
x = 2^4 \times 3^2 \times 5^2 \times 7^3
\]

Calculating the smallest prime factorization:
\[
x = 2^4 \times 3^2 \times 5^2 \times 7^3
\]
Evaluating \( x \):
The smallest third number is:
\[
\boxed{2^5 \times 3^5\times 5^2 \times 7^3} \text{ fully satisfies GCD and LCM }
\]

Therefore, \( x = 2^5 \times 3^5 \times 5^2 \times 7^3 \) indeed gives a unique smallest placement yielding \(144\) in GCD!
Calculating values gives:

Result- as:
\[
x = 2^5 \times 3^5 \times 5^2 \times 7^3
\]

Hence validated:
\[
\boxed{14400}
\] as valid third candidate!