Which of the following could be a value of k if 10^k and 12^k are fact

IanStewart. Dyamnn..

Didn't think that way.??

Posted from my mobile device

Hi,

I didn't understand what you meant by that. It will be really helpful if you can explain it in details.

Thank you

Posted from my mobile device

Hi,

How did we infer that 50! has two 2s and one 3.? And why is it important to find the max power of 5 of 50!.?

Thank you.

Posted from my mobile device

Nups1324

What he meant was if 10^16 is a factor of 50!, automatically 10^12, 10^13,10^14 and 10^15 would become factors of 50!

And since there can only be one answer straight away you can mark A.

Posted from my mobile device

Oh. Damn. I knew this rule but I couldn't see how I can use this.

That was smooth Ian. And thanks to you Harsh, I read it again and it clicked.

Can you explain me your working in detail? Harsh9676.

I really want to learn how you did it.

Thank you.

50! = 1*2*3*4*...*50. So 50! Has 2 twos and one 3.

We can also infer that the no of 2 s and no of 3 s in 50! Would be more than the no of 5s. This means that the no of 12s would be higher than no of 10s. So the limiting factor is 5.


So we need to find the no of 5 s in 50!.

Posted from my mobile device

I'm really sorry, but how 50! has only two 2s and one 3.

1*2*3*(2*2)*5*(2*3).....*50?

This part I understood, "We can also infer that the no of 2 s and no of 3 s in 50! Would be more than the no of 5s. This means that the no of 12s would be higher than no of 10s. So the limiting factor is 5."



Secondly, what is this formula that you used to find the max power of 5 in 50! in your original answer.?

[ Quotient of (50/5 )+ Quotient of (50/5^2) = 10 + 2 = 12 ]


Thank you.

Hi Nups1324

No Problem at all.

50! has obviously more than two 2's and one 3.

Two 2's and one 3 is the bare minimum (for 12 to be a factor of 50!).

Basically K should be equal to a number that satisfies the condition that 12^K and 10^K are factors of 50!. Since the the no of 2's and 3's (12 = 2^2 * 3) would be more than no of 5's (10 = 5 *2), we will have more 12's than 10's. So the limiting factor here is the no of 5's.

Regarding the formula: you can find it in this post.

https://gmatclub.com/forum/math-number- ... 88376.html

-Harsh

The question is asking you to find out the possible value of k such that \(10^k\) and \(12^k\) are factors of 50! or in other words the value of k such that 50! is perfectly divisible by \(10^k \)and \(12^k\).

So we can use the maximum power concept of a factorial to find the value of k.

\(10^k\) can be prime factorized as \(2^k \)* \(5^k\)

\(12^k\) can be prime factorized as \(2^k\) * \(2^k\) * \(3^k\)

Since the prime factors here are 2, 3, and 5, the maximum value of k depends upon the maximum power of the maximum prime (In this case the number of 5s).
So divide 50! by 5 until you cannot divide anymore and add up the quotients

\(\frac{50}{5} +\frac{50}{25^}\)
Adding the two quotients, we get 12. (Perform successive division)

So, the maximum value of k is 12.
(Remember that k = 12 is the maximum value k can take)
* Note : k can also be any value ≤ 12 which will still make \(10^k\) and \(12^k \)factors of 50!.

Option A is the answer.

Thanks,
Clifin J Francis

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.