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Bunuel
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 28 Oct 2021, 03:30
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If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
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D
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 28 Oct 2021, 04:00
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Bunuel
If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
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To find the highest power of 6 (=2*3) in 33!, we need to first find the highest power of 2 and 3 in 33! separately and then choose the smaller of the two powers.

Obviously the power will be smaller for 3. For 3, the power is:

33/3= 11
11/3=3.6 = 3 (integer part)
3/3= 1
So: 11+3+1= 15

So, for 33!, we have 6^15

We now need to repeat this for 22!
22/3=7.33 = 7 (integer)
7/3=2.33 = 2 (integer part)
So: 7+2= 9

So, for 22!, we have 6^9


Thus, for 33!/22!, we have: 6^(15-9) = 6^6

Answer D

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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 13 Sep 2022, 20:25
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sujoykrdatta
Bunuel
If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

To find the highest power of 6 (=2*3) in 33!, we need to first find the highest power of 2 and 3 in 33! separately and then choose the smaller of the two powers.

Obviously the power will be smaller for 3. For 3, the power is:

33/3= 11
11/3=3.6 = 3 (integer part)
3/3= 1
So: 11+3+1= 15

So, for 33!, we have 6^15

We now need to repeat this for 22!
22/3=7.33 = 7 (integer)
7/3=2.33 = 2 (integer part)
So: 7+2= 9

So, for 22!, we have 6^9


Thus, for 33!/22!, we have: 6^(15-9) = 6^6

Answer D

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Can you explain the logic as to how you derived the power? What property/theorem is this? sujoykrdatta Bunuel @expertgmat egmat
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 13 Sep 2022, 20:42
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Bunuel
If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

To find the highest power of 6 (=2*3) in 33!, we need to first find the highest power of 2 and 3 in 33! separately and then choose the smaller of the two powers.

Obviously the power will be smaller for 3. For 3, the power is:

33/3= 11
11/3=3.6 = 3 (integer part)
3/3= 1
So: 11+3+1= 15

So, for 33!, we have 6^15

We now need to repeat this for 22!
22/3=7.33 = 7 (integer)
7/3=2.33 = 2 (integer part)
So: 7+2= 9

So, for 22!, we have 6^9


Thus, for 33!/22!, we have: 6^(15-9) = 6^6

Answer D

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Can you explain the logic as to how you derived the power? What property/theorem is this? sujoykrdatta Bunuel @expertgmat egmat
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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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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 13 Sep 2022, 21:12
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Bunuel
If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

To find the highest power of 6 (=2*3) in 33!, we need to first find the highest power of 2 and 3 in 33! separately and then choose the smaller of the two powers.

Obviously the power will be smaller for 3. For 3, the power is:

33/3= 11
11/3=3.6 = 3 (integer part)
3/3= 1
So: 11+3+1= 15

So, for 33!, we have 6^15

We now need to repeat this for 22!
22/3=7.33 = 7 (integer)
7/3=2.33 = 2 (integer part)
So: 7+2= 9

So, for 22!, we have 6^9


Thus, for 33!/22!, we have: 6^(15-9) = 6^6

Answer D

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Can you explain the logic as to how you derived the power? What property/theorem is this? sujoykrdatta Bunuel @expertgmat egmat
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The first 33 numbers have 11 multiples of 3
Those 11 multiples have 11/3 = 3 multiples of 3 more (ie 9)
These 3 have 3/3 = 1 multiple of 3 more (ie 27)

So 11+3+1=15

Hope this helps
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 13 Sep 2022, 21:50
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Asked: If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?

Highest power of 3 in 33! = 11 + 3 + 1 = 15
Highest power of 3 in 22! = 7 + 2 = 9
Highest power of 3 in 33!/22! = 15 - 9 = 6

Highest power of 2 in 33! = 16 + 8 + 4 + 2 + 1 = 31
Highest power of 2 in 22! = 11 + 5 + 2 + 1 = 19
Highest power of 2 in 33!/22! = 31 - 19 = 12

Maximum possible value of k = min (6,12) = 6

IMO D
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 15 Sep 2022, 05:40
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Bunuel
If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

To find the highest power of 6 (=2*3) in 33!, we need to first find the highest power of 2 and 3 in 33! separately and then choose the smaller of the two powers.

Obviously the power will be smaller for 3. For 3, the power is:

33/3= 11
11/3=3.6 = 3 (integer part)
3/3= 1
So: 11+3+1= 15

So, for 33!, we have 6^15

We now need to repeat this for 22!
22/3=7.33 = 7 (integer)
7/3=2.33 = 2 (integer part)
So: 7+2= 9

So, for 22!, we have 6^9


Thus, for 33!/22!, we have: 6^(15-9) = 6^6

Answer D

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Can you explain the logic as to how you derived the power? What property/theorem is this? sujoykrdatta Bunuel @expertgmat egmat
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PART 1: Understanding the Logic
Let us first try to understand the logic using a simpler example:
  • Suppose we need to find the maximum power of 2 in 10!
    • Observe that 10! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 is a product of 10 integers, of which five are even.
      • These are 2, 4, 6, 8, and 10. Each of these has at least one power of 2.
      • So, the product 2 × 4 × 6 × 8 × 10 will have at least these FIVE 2s.
        • Let’s split 2, 4, 6, 8, and 10 to see the five 2s we get:
          • = (2×1) × (2×2) × (2×3) × (2×4) × (2×5)
          • = \(2^5\) × (1 × 2 × 3 × 4 × 5)
          • = \(2^5\) × 5! --------------------(1)
    • Now, we need to check the power of 2 in this 5! = 1 × 2 × 3 × 4 × 5.
      • Again, we have two multiples of 2 in 5!.
      • These two multiples, 2 and 4, will contribute at least one 2 each.
        • Let’s split 2 and 4 to see the two 2s we get:
          • = (2 × 1) × (2 × 2)
          • = \(2^2\) × (1 × 2) --------------------(2)
    • Finally, we will check the number of 2s in 2!
      • 2! = \(2^1\). This has just ONE 2. --------------------(3)
    • Overall, we get a total of EIGHT 2s: \(2^5 × 2^2 × 2^1 = 2^8\)
      • This implies the maximum power of 2 in 10! is 8.


PART 2: Deriving the formula
Now, if you observe this carefully, you will see that:
  1. First, we find the number of multiples of 2 in 10!.
    1. This number is equal to the quotient on dividing 10 by 2. It can be represented as [10/2].
    2. So, [10/2] = 5.
  2. Then, we use this 5 and find the multiples of 2 in 5!
    1. This number is equal to the quotient on dividing 5 by 2. It can be represented as [5/2].
    2. So, [5/2] = 2.
  3. Then, finally, we use this 2 and find the number of multiples of 2 in 2!.
    1. This number is equal to the quotient on dividing 2 by 2. It can be represented as [2/2].
    2. So, [2/2] = 1.

  • Overall, the total number of 2s in 10! = \([\frac{10}{2}]+[\frac{5}{2}]+[\frac{2}{2}] = 5 + 2 + 1 = 8\)
  • This is not a co-incidence. We actually have this result that helps us find the power of any prime number in the prime factored form of a factorial, say N!.
    • The result is as follows:
      • Power of p in \(N! = [\frac{N}{p}]+[\frac{N}{p^2} ]+[\frac{N}{p^3 }]+[\frac{N}{p^4 }]+⋯\)
        • Here, the square brackets denote the quotient on dividing N by p.


PART 3: Cement your understanding
Before we conclude, let us try to solve another example and this time, we will directly use this result.
  • Example: Find the numbers of 3s in 20!
    • For the above formula, we first calculate the following:
      • \([\frac{N}{p}]= [\frac{20}{3}] = 6\)
      • \([\frac{N}{p^2} ]= [\frac{20}{3^2} ]=2= [\frac{6}{3}]\)
      • \([\frac{N}{p^3 }]= [\frac{20}{3^3} ]=0= [\frac{2}{3}]\)
        • We should stop here, as for every term after this we will get 0 as quotient.
        • So, the maximum power of 3 in 20! = 6 + 2 = 8

Hope this helps!
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 18 Oct 2023, 05:05
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took me 1.08 mins
33!/22!=33*32*31*30*29*28*27*26*25*24*23*22!/22!
=33*32*31*30*29*28*27*26*25*24*23

6=2*3
33 =3*11, one 3
30=3*10, one 3
27=3*3*3, three 3
24=3*8, one 3
so there are 6 threes

so K can go to the power of 6
IMO D
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu 04 Apr 2026, 04:40
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Have to take the limiting factor = 3.
33! has 3^15 as its max factor of powers of 3.
22! has 3^9 as its max factor of powers of 3.
Dividing the 2 we see we can have 15-9=6 as the value of k.

Answer: Option D
Bunuel
If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
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