Which of the following is NOT a factor of the product of the

Question

Which of the following is NOT a factor of the product of the first 50 positive multiples of 4 ?

A.  17^2 
B.  11^4 
C.  7^6 
D.  47^{12} 
E.  2^{124}

blueseas · Accepted Answer

product of the first 50 positive multiples of 4 = 4*8*12*16*......200 =4^50*(1*2*3......49*50) = 2^100(FACTORIAL 50)

lets check options:

A.  17^2  ==>we need two 17 ==>that can be found in FACTORIAL 50 (17,34)=>factor

B.  11^4  ==>we need four 11==> that can be found in FACTORIAL 50 (11,22,33,44)=>factor

C.  7^6  ==>we need six 7 ==>that can be found in FACTORIAL 50(7,14,21,28,35,42)=>factor

D.  47^{12}  ==> not a factor since we 47 is a prime and is only 1 time present in factorial 50

E.  2^{124} ==>This can be found as we have 2 ^100 and in factorial 50 we have 25 even numbers=>factor

hence D

Asifpirlo · Answer

4^50×(1.2.3.4.5.6.7..............................50)

= 4^50 (1.2.3.4.................50)

17^2 = 17 ×17 (we have 17 and 34) so this is a factor.

11× 11× 11× 11 (we have 11,22,33,44) so this is a factor.

7 × 7× 7× 7 ×7 ×7 (we have 7,14,21,28,35,42) so this is a factor.

2^124 = (4)^62 . Now 4^50 / 4^62 = 1/4^12 = 1/(4 . 4 . 4. 4. 4. 4. 4. 4. 4. 4. 4. 4) we have (4,8,12,16,20,24,28,32,36,40,44,48) . so this is also a factor.

These left only one as the guilty and it's 47^12

stonecold · Answer

Firstly the product of first positive 50 multiples of 4=>
4*1*4*2*4*3*4*4*4*5.....4*50=> 4^50*50!
Secondly notice Option D is 47^2 and 47 is a prime number.
Number of 47's in 4^50*50!=> one
Hence 47^2 will never be its factor.

Hence D.

Shiv2016 · Answer

Hi  Bunuel  
Can you please give your explanation for this question..All the 50 multiples of 4 would be even, then how come 17 is a factor of this product?

Bunuel · Answer

My solution would be the same as this one: https://gmatclub.com/forum/which-of-the ... l#p1257006

Bunuel · Answer

Also, are you saying that even number cannot have an odd factor? It's not true. For example, 4*17 = 68 is even and is a multiple of 17.

ScottTargetTestPrep · Answer

Since the first positive multiple of 4 is 4 and the 50th is 200, we see that in the product of the numbers of:

4 x 8 x 12 x 16…..x 196 x 200, we won’t have 12 factors of 47, and thus, 47^12 is not a factor of the first 50 multiples of 4. (Note: 47 is prime, and in the product 4 x 8 x 12 x 16…..x 196 x 200, there is only one factor of 47, which is in 188.)

Answer: D

Rebaz · Answer

Hi,

Can some please explain how can the below quoted portion

Equal to

Without skipping any simplification steps?

Thanks in advance!

dave13 · Answer

hey there Rebaz for Example Factorial 50! means 1*2*3*4*....50 since we need to find product of first 50 multiples of 4 so basically you multiply 4 by consecutive first 50 numbers (4*1, 4*2, 4*3, 4*4 ....* 4*50) which is the same as 4^{50}*(1*2*3......49*50) or 2^{100} *50!

hi Bunuel , this question is tagged both as kaplan and OG, i wonder from which source is the question

Rebaz · Answer

dave13

Thank you for your clear explanation!

SPFF · Answer

4 x 8 x 12 x...........x200 = 4^50(50!), now looking on 50! i.e (1 x 2 x 4 x.....x 50) separately, which comprises of two 17s (17, 34 (17 x2), four 11s (11, 22, 33,44), eight 7s (7,14,21,28,35,42 and 49 (49 = 7 x7) and on a similar manner number of twos in 50! would be (25 (2s)+ 12 (4s)+ 6 (8s)+ 3 (16s)+ 1(32s) = 47 ) + 2s in in 4^50 (=2^100 = 100), therefore total of 147. But for 47 there is only one factor of 47 (<50) which is 47.

Any number beyond the accounted #s (on power) is not a factor. Be it either 17^3 or 11^5 or 7^9 or 2^148 or even 47^2. Hence D.

Any number beyond the accounted #s (on power) is not a factor. Be it either 17^3 or 11^5 or 7^9 or 2^148 or even 47^2. Hence D.

hellothere88 · Answer

I thought it mentioned first 50 MULTIPLES of 4?not 50!?so its 4, 8, 12, ..., 200?

GMATBUDDING · Answer

Can anyone help me in explaining why option E is correct ?

Bunuel · Answer

The product of the first 50 positive multiples of 4 is  4*8*12*...*200 = 4^{50}(1*2*3*...*50) = 2^{100}*50! .

To find the power of 2 in the prime factorization of 50!, we use the formula: 50/2 + 50/4 + 50/8 + 50/16 + 50/32 = 25 + 12 + 6 + 3 + 1 = 47 (only the quotients of the divisions are considered, for example, 50/16 = 3). Thus,  2^{100}*50!=2^{100}*2^{47}*something=2^{147}*something , which is divisible by 2^124.

For theory check:  Number Properties

For questions check the following topics from our  Special Questions Directory :

12.  Trailing Zeros 
13.  Power of a number in a factorial

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

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