If f(n) is the product of n consecutive integers from

Question

If  f(n)  is the product of  n  consecutive integers from  1  to  N , then what is the number of odd factors of f(10) .

A) 30
B) 60
C) 120
D) 256
E) 512

e-GMAT Question:

This is  Question 11 of  The e-GMAT Number Properties Marathon

Go to   the next level of the Marathon

EgmatQuantExpert · Accepted Answer

Solution:

We are given:
  F(n)= 1×2×3×4×5×……….×n 
We need to find the number of odd factors of  f(10) .
Thus, we need to find the value of  f(10).  Then, we need to write  f(10)  in its prime factorized form. After which we can calculate the value of odd factors of f(10) .
Number of Odd factors  f(10) 
  f(10)= 1×2×3×4×5×6×7×8×9×10  
  f(10)= 1×2×3×(2×2)×5×(2*3)×7×(2*2*2)×(3*3)×(2*5)  
  f(10)= 2^8 * 3^4 * 5^2 *7  
Odd factors of  f(10) = (power of 3+1)*(power of 5+1)* (power of 7+1)
Odd factors of  f(10)=5*3*2 
Odd factors of  f(10)=30

MikeScarn · Answer

The principle this question is testing is  Finding the Number of Factors of an Integer

Number of  odd  factors of 10!

10!=1x2x3x4x5x6x7x8x9x10.
10!=2x3x( 2^{2} )x5x(2x3)x7x( 2^{3} )x( 3^{2} )x(5x2)
10!= 2^{8}  3^{4}  5^{2} 7

Take the exponents of the odd factors. Add 1 to each of them and multiply them by each other.
(4+1)(2+1)(1+1)=30

Answer = A

Hatakekakashi · Answer

10! = 1 x 2 x 3 x 2^2 x 5 x 2 x 3 x 7 x 2^3 x 3^2 x 2 x 5
10! = 2^8 x 3^4 x 5^2 x 7

factors = 9 x 5 x 3 x 2

odd factors = 5x3x2

(A) 30 imo

EgmatQuantExpert · Answer

Solution:

We are given:
  f(n)= n! 
Thus we can write:
   f(10)= 10! 
  10! = 1×2×3×4×5×6×7×8×9×10 
  10!= 2^8 * 3^4 * 5^2 *7  
Total factors of  10!= 9*5*3*2=270 
Total factors = Even factors + Odd factors
Odd factors= Total factors – Even factors
Thus, if we can find the even factors of  10!  then we can find the odd factors of  10! .
  10!= 2*(2^7 * 3^4 * 5^2 *7) 
We can say that all the factors of  2^7 * 3^4 * 5^2 *7  after multiplying by  2  becomes a factor of  10! . 
We also know that any number, which is multiple of  2 , is an even number. 
Thus, 
   Total number of factors of 2^7 * 3^4 * 5^2 *7  = Even number of factors of  10!.  
 Even factors of  10! = Total factors of  2^7 * 3^4 * 5^2 *7 
 Even factors of  10! =  8*5*3*2 
 Even factors of  10! = 240  
Therefore, Odd factors= Total factors – Even factors
 Odd factors=  270-240 
 Odd factors= 30 
Answer: Option  A

GMATBusters · Answer

HI I feel question would have been more interesting if there is an option 270 in the answer choices. If someone fail to read odd factors (as I did

), one would mark 270 as answer. Fortunately, answer choices saved me. Regards

kabirchaudhry92 · Answer

Hello!

I have a doubt here.. Why is 1 not included in the odd factors ? 
Sorry for such a silly question but isn't that supposed to be counted too ?

It would be really helpful if you could provide your insights on the same.

Thanks in advance

ganand · Answer

Hi  kabirchaudhry92 ,

This is a good question. Let me explain with some small examples.

Q1. Find the number of factors of 25.

Step 1: Do the prime factorization. 25 = 5*5 =  5^2

Step 2: Number of factors = (2+1) = 3.

Why have we added 1? By adding "1" we ensure that we count 1 as one of the factors.

Factors of 25:  5^0 = 1, 5^1 = 5,  and  5^2 = 25.  Hence, the answer is 3.

Q2. Find the number of factors of 75.

Step 1: Prime factorization = 3*5*5 =  3^1 * 5^2

Step 2: Number of factors = (1+1) * (2+1) = 6.

Factors of 75:
    5^0 * 3^0 = 1  
  5^0 * 3^1 = 3  
  5^1 * 3^0 = 5  
  5^1 * 3^1 = 15  
  5^2 * 3^0 = 25  
  5^2 * 3^1 = 75   .

Hence, 1 is already included in the given solution.

I hope this helps.

Thank you.

pappal · Answer

f(10)=!10=2^8*3^4*5^2*7^1
 for total no of odd factors we must not consider any even factor or discard all the powers of 2 or simply find the no. of factors of 3^4*5^2*7^1
 so total no. of odd factors of !10=(4+1)*(2+1)*(1+1)=30

If f(n) is the product of n consecutive integers from

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