P denotes the product of all the even natural numbers upto F, and Q

Given in question,

P denotes the product of all the even natural numbers upto F, and Q denotes the product of all the odd natural numbers upto F.

This means product of P and Q gives the product of all natural numbers upto F.

So product of all natural numbers upto F is a multiple of 3213.

3213 = 3*3*3*7*17

From above product of prime factors we can conclude that the product of all natural numbers till 17 results in a multiple of 3213.

OA: (D)

Solution

Given:

• P = product of all the even natural numbers upto F.
• Q = product of all the odd natural numbers upto F.
• The product of P and Q is a multiple of 3213.

To find:

• The minimum value of F.

Approach and Working:
As per the definition of P and Q, P is the product of all even natural numbers upto F and Q is the product of all odd natural numbers upto F.

• Therefore, P * Q = product of all the natural numbers upto F = Factorial of F.
• So, as per the given question, F! is a multiple of 3213.

Now, if we express 3213 in terms of its prime factors, we get \(3213 = 3^3 * 7 * 17\)

Therefore, in F!, we will have \(3^3, 7\) and \(17\) at least.

So, the minimum value of F must be 17.

Hence, the correct answer is option D.

Answer: D
_________________

The product of P and Q is a multiple of 3213. This means P*Q must have all factors of 3213.

jojo95 - To factorise a large number, it helps to know multiplication tables till 20.
We can easily see that 3 is a factor of 3213 (since 3+2+1+3 = 9 is divisible by 3). But this also shows us that 3213 is divisible by 9.

Hence first split is
3213 = 9*357

Now 357 is divisible by 7 because 35 is a multiple of 7 and then we have 7. So second split is
3213 = 9*7*51

Finally, we know that 17*3 = 51
3213= 9*7*17*3

Since 17 is the largest prime factor of 3213 and the product P*Q has 17 as a factor, it means F must be at least 17. Only then can P*Q have 17 because we cannot make 17 using any other factors.
P = 2*4*6*8*...*16
Q = 1*3*5*7*...*17

Answer (D)
_________________

sebastianojara & ZulfiquarA - I will take a shot to explain. Let me know if it helps?


Yes, \(PQ\) will be even. The question does not state that \(PQ\) equals \(3,213\) (odd integer) rather it states that \(PQ\) is a multiple of \(3,213\)

Let's assume the \(PQ = 6\) (even). Here we can also say that \(PQ\) is a multiple of odd integer \(3\) even though \(PQ\) is even


1. \(3,213 = 3^3 * 7 * 17\)

2. We are told that \(PQ\) is a multiple of \(3,213\) i.e. \(3,213k\) where \(k\) is a positive integer. So \(PQ = 3,213 * 2\) or \(3,213 * 4\) or \(3,213 * 6\) and so on...

3. Let's take the smallest value \(3,213 * 2\) for understanding purpose

4. \(P * Q\) \(=\) \((2 * 4 * 6 * ... * F) * (1 * 3 * 5 * ... * F)\) \(=\) Some even number

5. Now for \(PQ\) to be \(3,213 * 2\) it must also contain all prime factors and factors of \(3,213\). We need \(PQ\) to contain three \(3\)'s, one \(7\), one \(17\) and one \(2\)

6. We first need three \(3\)'s. Substitute \(F = 9\) in pt #4 and you will see that the product contains three \(3\)'s, one from \(3\) itself and other two from \(9\). You will not find the three \(3\)'s if \(F < 9\)

7. Now, we also need a \(7\) and a \(2\). Again \(F = 9\) contains a \(7\) and a \(2\) inside it

8. Next, we need a \(17\). Tell me what value of \(F\) gives a \(17\)? Only \(F >= 17\) will give you a \(17\) since \(17\) also happens to be prime number and you will not find it if \(F < 17\)

9. To summarize, \(F = 17\) is the minimum value that contains three \(3\)'s, one \(7\), one \(17\) and one \(2\) because if \(F < 17\) then \(PQ\) cannot be a multiple of \(3,213\)

Hope it helps!

factorize 3213 ; 3^3*7*17
min value of F has to be 17 ; to be a multiple of 3213
IMO D

HI! Ive been struggling with this question. Shouldnt the product of Q and P be necessary an even number? And why should 17 be the minimum value if we also have 7?

What is the easiest way to find the prime factors of such a large number? I took a very long time to solve this problem and it probably shouldn't have taken me this long...

Any tips or tricks would be much appreciated! Thank you in advance!

Hey,
There are no tricks to prime factorisation or at least I'm not aware of any. Irrespective, the given number is not hard to prime factorise.
To tell you the truth, when I started solving the question myself, my first thought on seeing the number was there is no way GMAT will ask me to factorise this. There has to be a better way.
After wasting a minute and failing to come up with an alternate method, I did what was needed. PRIME FACTORISATION! and to my surprise it was really easy. The only thing holding me back was my fear/thought.

I don't know why you took a really long time to solve this. If you were stuck on where to start factorising - JUST PICK A NUMBER. Start somewhere, move on, repeat!

Hope I helped you in some way.

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.
_________________