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. $6480÷10=648$.
2. $648÷9=72$, 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. $7200÷100=72$, 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: $$min(4,a)=4\Rightarrow a\ge 4$$ $$min(4,b)=2\Rightarrow b\ge 2$$ $$c\text{ must be }\ge 0$$ $$d\text{ may be }0\text{ (since it isn't part of previous numbers)}$$

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:

• $max(4,a)=5\Rightarrow a=5$
• $max(4,b)=5\Rightarrow b=5$
• $max(1,c)=2\Rightarrow c=2$
• $max(0,d)=3\Rightarrow d=3$

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: $$\boxed{{\displaystyle 2^5\times 3^5\times 5^2\times 7^3}}\text{ fully satisfies GCD and LCM }$$

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: $$\boxed{{\displaystyle 14400}}$$ as valid third candidate!