-1
$\begingroup$

What is the greatest common divisor of $2^4\cdot3^4\cdot25\cdot7$ and $2\cdot12^2\cdot15$?

I know how to find the GCD of a problem like this, but I didn't know what to do since $12$ is not a prime number. Should I say $12^2 = 2^2\cdot2^2\cdot3^2$?

$\endgroup$
  • $\begingroup$ Hint: $12^2 = (2^2*3)^2 = 2^4*3^2$ $\endgroup$ – AnalyticHarmony Sep 18 '17 at 2:22
  • $\begingroup$ that's a start. $12^2 =2^2*2^2*3^2=2^4*3^2$. And $15=3*5$ so $12^2*15=2^4*3^2*3*5 = 2^4*3^3*5$. Basically the first step is to rewrite everything as its unique prime factorization. $2^4*3^4*25*7 = 2^4*3^4*5^2*7$ and $2*12^2*15 = 2*2^4*3^2*3*5=2^4*3^3*5$. Then... you know what to do. $\endgroup$ – fleablood Sep 18 '17 at 3:49
1
$\begingroup$

The first one equal to $2^4*3^4*5^2*7$

The second one equals to $2^5*3^3*5$

Therefore the GCD is $2^4*3^3*5=2160$

$\endgroup$
1
$\begingroup$

Yes, that is precisely what you should be doing.

In short, when you want to find the gcd of two numbers directly, you must first represent each one as a product of prime powers, before matching powers and primes.

So in our case, we can write: $$ 2^4 \times 3^4 \times \color{blue}{25} \times 7 = 2^4 \times 3^4 \times \color{blue}{5^2} \times 7 \\ 2 \times \color{red}{12^2} \times \color{green}{15} = 2 \times \color{red}{2^2 \times 2^2 \times 3^2} \times \color{green}{3 \times 5} = 2^5 \times 3^3 \times 5 $$

where, I highlight by color the terms that are expanded on both the LHS and RHS.

Now, we are permitted to compare powers directly. This gives us the answer as $2^4 \times 3^3 \times 5$.

$\endgroup$
  • $\begingroup$ Nicely explained and nice use of colors. The OP should note when you express a number and a product of prime powers, there will always be exactly one unique way to do it. So doing this will always be a failsafe that will work. $\endgroup$ – fleablood Sep 18 '17 at 3:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.