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!