For each integer n 1, let g(n) equal the total number of

Question

For each integer  n > 1 , let  g(n)  equal the total number of (not necessarily distinct) prime factors of  n . For example,  g(6) =  2 ,  g(7 ) =  1 , and  g(8) = 3 . If N =  (10!)^{100} , what is  g(g(N)) ?

A. 6
B. 10
C. 15
D. 50
E. 100

This is a  PS Butler Question

Check the links to other Butler Projects:

Data Sufficiency Butler  
  Critical Reasoning Butler  
  Reading Comprehension Butler  
  Data Insights Butler

preetisehrawat · Accepted Answer

Step 1: Prime Factorization of N= (10!)^100

10! =10*9*8*7*6*5*4*3*2*1
 = 2*5*3*3*2*2*2*7*2*3*5*2*2*3*2*1
 = 2^8 * 3^4 * 5*2 * 7

Thus, (10!)^100 = ( 2^8 * 3^4 * 5*2 * 7)^100
 = 2^800 * 3^400 * 5^200 * 7^100 
Therefore, g(N) = 800+400+200+100
 = 1500

g(g(N)) = g(1500)
 
 1500 = 2^2 *3 * 5^3

g(1500)= 6
Hence,the answer is 6.

satish_sahoo · Answer

Time taking.

As per question to calculate prime factors of  10! 
2 has 1 prime factor
3 has 1
4 has 2
5 has 1
6 has 2
7 has 1
8 has 3
9 has 2
10 has 2

Total  15  prime factors, now  10!  has to be multiplied  100 times so we will get  1500 prime factors.

g(1500)  = prime factorization of  15  is  2,2,5,5,5,3 . Total = 6

I think A

amitarya · Answer

I want clarifications regarding factorisation of a number- 
Suppose the number is 450
Prime factorisation of 450 is
450 = 2^1 * 3^2 * 5^2
Then, as per my understanding

i) Nos. of Prime Factors = 1+2+2 = 5 (i.e. sum of power of all prime factors)

ii) Nos. of Unique prime factors = 3 (i.e., all unique prime factors- 2,3 & 5)

iii) Nos. of Factors = 2*3*3 = 18 (this includes all even and odd factors including 1 and number itself)

Please confirm, if my understanding is correct.

Bunuel

karishma

Pls help..

rRathi99 · Answer

Yup, all three things you listed out here are correct. This question is however asking for all prime factors, regardless of repetition. Note, g(8) = 3. Prime Factorisation of 8 = 2*2*2.

accusantiumsit · Answer

how do I find prime factors quickly? like the answer makes sense to me but i don`t know how to make factorization. please spmeone explain

Krunaal · Answer

I would recommend you go through Prime nos., Factors, and Divisibility here https://gmatclub.com/forum/math-number- ... 88376.html One way to get prime factors quickly is to have a good grasp on divisibility rules, eg. if you want to do prime factorization of 1323, you easily know that it can be divided by 9, if you know the divisibility rule for 9 is if sum of all digits is a multiple of 9 then that no. is divisible by 9 so 1+3+2+3 = 9, so number has 3^2 as a factor and the quotient can be divided further to find other prime factors, also concept of primality (explained in the above link) is good to know when none of the divisibility rules look applicable.

For each integer n > 1, let g(n) equal the total number of

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions