Let n be the least positive integer that has exactly 5 different prime

Question

Let n be the least positive integer that has exactly 5 different prime factors. What is the greatest value of k such that  2^k  divides  \frac{12!}{n} ?

A. 7
B. 8 
C. 9
D. 10
E. 11

gmatophobia · Accepted Answer

As n is the least positive integer with exactly 5 different prime factors,

n = 2 * 3 * 5 * 7 * 11

Note: All the prime factors except 2 are odd.

Let's find the power of 2 in 12!

Power of 2 in 12! =  \frac{12}{2} + \frac{12}{4 }+ \frac{12}{8} + \frac{12}{16} = 6 + 3 + 1 + 0 = 10

As  n  consists of only a single power of 2 (i.e.  2^1 ) ,  \frac{12!}{n} = \frac{2^{10} * 	ext{some value}}{2^1 * 	ext{some other value}} = 2^9 * 	ext{some value}

Hence, the greatest value of k such that  2^k  divides  \frac{12!}{n}  is 9.

Option C

naman.goswami · Answer

n has to be the multiple of first five prime numbers. 
n = 2*3*5*7*11

Total number of 2s in 12! can be calculated by adding the coefficient when 12 is divided by 2^n
12/2= 6
12/4= 3
12/8 =1

Number of times 2 is multiplied in 12! = 6+3+1 = 10 
so 12!/n has 9 times 2

Hence k=9

Kinshook · Answer

jcmsf98 
Given: Let n be the least positive integer that has exactly 5 different prime factors.
Asked: What is the greatest value of k such that  2^k  divides  \frac{12!}{n} ?
n = 2*3*5*7*11

Greatest power of 2 in 12! = 6 + 3 + 1 = 10
Greatest power of 2 in 12!/n = 10-1 = 9 ; Since n has exactly 1 power of 2 as its factor

IMO C

ShilpiAgnihotrii · Answer

n= 2*3*5*7*11

12!/n will result in = 12*10*9*8*6*4

Prime factorization of 12*10*9*8*6*4 = 2^9*3^4*5

So highest power of k is 9. 
Ans: C

satish_sahoo · Answer

So, n=2*3*5*7*11

When we divide 12! by n, we remain with 12*10*9*8*6*4.

12 has Two 2s
10 has One 2
9 has zero 2
8 has Three 2s
6 has One 2
4 has Two 2s

Hence total 2s we got is 9. 
Ans - C.

ablatt4 · Answer

This question is a really easy question that is really hard to decipher.

What we know is we are dividing 12! by n which is an integer with 5 distinct prime factors, and we are asked for the great value of 2^k that factors into the term we must solve for.

The trick here, their is only 1 prime that is even, which is 2. Knowing this we can reevaluate what is being asked, if we divide 12! by n we can only factor out a maximum of 1 2 from this term. If K represents the maximum amount of 2s we can continue to factor out we can solve this by figuring out how many 2s are in 12! and subtracting 1.

9

Jujustrollss · Answer

n is the smallest number with 5 prime factors- ie; 2x3x5x7x11

12!= 2x3x4x5x6x7x8x9x10x11x12
Now we divide it by “n”, we have

4x6x8x9x10x12
= 2^5(2x3x4x5x6) x 9
= 2^5(6!) x 9

Number of powers of 2 that divide 6! Are 4
No 2s in 9
2^5 we have factorized out

Hence, 2^5 x 2^4
So the largest power of 2 to divide 12!/n is 5+4=9

Please let me know if this logic is correct

egmat · Answer

Jujustrollss

Great job identifying  n = 2 	imes 3 	imes 5 	imes 7 	imes 11 !  That's absolutely correct - the smallest number with exactly 5 different prime factors.

Process Diagnosis: 
Your cancellation approach has a critical error. When you wrote  \frac{12!}{n} = 4 	imes 6 	imes 8 	imes 9 	imes 10 	imes 12 , you essentially "removed"  2, 3, 5, 7, 11  from the factorial. However,  12!  contains  multiple  instances of some primes (like multiple  2 s and  3 s), so you can't just remove them from the list of factors.

Correct Approach: 
To find the highest power of  2  dividing  \frac{12!}{n} , use this principle:
 	ext{Power of 2 in } \frac{12!}{n} = 	ext{Power of 2 in } 12! - 	ext{Power of 2 in } n

Step 1:  Find the power of  2  in  12!  using Legendre's formula:
 \left\lfloor\frac{12}{2}ightfloor + \left\lfloor\frac{12}{4}ightfloor + \left\lfloor\frac{12}{8}ightfloor + \left\lfloor\frac{12}{16}ightfloor + ... 
 = 6 + 3 + 1 + 0 = 10

So  2^{10}  divides  12!

Step 2:  Power of  2  in  n = 1  (since  n = 2^1 	imes 3 	imes 5 	imes 7 	imes 11 )

Step 3:  Therefore, power of  2  in  \frac{12!}{n} = 10 - 1 = 9

Why Your Answer Was Still Correct: 
Interestingly, you got  9  through a different path! Your calculation  2^5 	imes 2^4 = 2^9  happened to work out, but the intermediate steps weren't accurate.

Strategic Insight - Factorial Prime Power Pattern:

When you see "highest power of prime  p  dividing  \frac{n!}{k} :
1. Always calculate powers separately using Legendre's formula for the factorial
2. Find the prime factorization of  k 
3. Subtract: (power in factorial) - (power in denominator)
Never try to "cancel" factors directly from the factorial expansion.

pikachu9 · Answer

1,2,3,5,7- 5 different prime factors for n . so by removing them from 12! we are left with 4,6,8,9,10,11,12
4=2^2
6=2*3
8=2^3
9=3^2
10=2*5
11=11*1
12=2^2*3

so total value of k= 9 
Is this method correct?

LushPear · Answer

im not following how we know that n only consists of a single power of two... is it because we know n has the least amount of 5 distinct primes so n already has one 2 in it? and thats why we subtract it??

Bunuel · Answer

Yes. Since n is the  smallest  positive integer with exactly 5 different prime factors, it must be 2 * 3 * 5 * 7 * 11, using the five smallest distinct primes once each.

So n contains only one factor of 2, which is why we subtract just 1 from the exponent of 2 in 12!.

Let n be the least positive integer that has exactly 5 different prime

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