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

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. What is the minimum value F can have, if the product of P and Q is a multiple of 3213?

A.

B.

C.

D.

E.

A.  7
 B.  9
 C.  13
 D.  17
 E.  27

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-NP-e1533292144892.png

madgmat19 · Accepted Answer

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)

Archit3110 · Answer

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

EgmatQuantExpert · Answer

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

jojo95 · Answer

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!

Karmesh · Answer

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.

sebastianojara · Answer

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?

KarishmaB · Answer

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)

Pritishd · Answer

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!

DonBosco7 · Answer

We can also work around this question using answer choices, though it might take a few more seconds that the prime factorization method described above.

given: P = 2x4x6x8........xF && Q = 1x3x5x7..........xF

PxQ is a multiple of 3213 (i.e/ 3213 x something = P x Q)

also notice P x Q = 1x2x3x4x5x6x7............xF 
which is F!
hence we can write this as

something (say "k") x 3213 = F!
k x 3213 = F!
k = F!/3213
if you try any option choice for F, only 17 will fit in, as in when you put F! = 17! then only 3213 will completely divide the equation.
D

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

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