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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
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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
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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.
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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
jcmsf98
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

­
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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
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Jujustrollss
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
Jujustrollss

Great job identifying \(n = 2 \times 3 \times 5 \times 7 \times 11\)! 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 \(\frac{12!}{n} = 4 \times 6 \times 8 \times 9 \times 10 \times 12\), 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:
\(\text{Power of 2 in } \frac{12!}{n} = \text{Power of 2 in } 12! - \text{Power of 2 in } n\)

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

So \(2^{10}\) divides \(12!\)

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

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 \(2^5 \times 2^4 = 2^9\) 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.
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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?

jcmsf98
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

­
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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??
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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??
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!.
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