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# What is the greatest value of x such that 8^x is a factor of 16! ?

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Math Expert
Joined: 02 Sep 2009
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What is the greatest value of x such that 8^x is a factor of 16! ? [#permalink]

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14 Dec 2016, 04:57
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What is the greatest value of x such that 8^x is a factor of 16! ?

A. 2
B. 3
C. 5
D. 6
E. 8
[Reveal] Spoiler: OA

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Re: What is the greatest value of x such that 8^x is a factor of 16! ? [#permalink]

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14 Dec 2016, 05:39
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First we'll find power of $$2$$ in $$16!$$

$$[\frac{16}{2}] + [\frac{16}{2^2}] + [\frac{16}{2^3}] + [\frac{16}{2^4}] = 8 + 4 + 2 + 1 = 15$$

and we have:

$$2^{15} = (2^3)^5 = 8^5$$

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Re: What is the greatest value of x such that 8^x is a factor of 16! ? [#permalink]

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14 Dec 2016, 07:02
Bunuel wrote:
What is the greatest value of x such that 8^x is a factor of 16! ?

A. 2
B. 3
C. 5
D. 6
E. 8

8 = 2^3

Highest power of 2 in 16! is 15

16/2 = 8
8/2 = 4
4/2 = 2
2/2 = 1

So, The highest power of 8 in 16! will be 15/3 = 5

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Re: What is the greatest value of x such that 8^x is a factor of 16! ? [#permalink]

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15 Dec 2016, 17:15
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Expert's post
Bunuel wrote:
What is the greatest value of x such that 8^x is a factor of 16! ?

A. 2
B. 3
C. 5
D. 6
E. 8

Since 8 = 2^3, we are actually trying to determine the greatest value of x such that 2^(3x) is a factor of 16!.

Let’s first determine the number of factors of 2 within 18!. To do that, we can use the following shortcut in which we divide 18 by 2, and then divide the quotient of 18/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

16/2 = 8

8/2 = 4

4/2 = 2

2/2 = 1

Since 1/2 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 2 within 16!.

Thus, there are 8 + 4 + 2 + 1 = 15 factors of 2 within 16!

However, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 8. We see that 15 factors of 2 will produce 5 factors of 8.

Note: To clarify the final answer, note that the 16 factors of 2 can be expressed as 2^16. We now must break this number 2^16 into as many factors of 8 as possible; thus, we will have

2^16 = 2^3 x 2^3 x 2^3 x 2^3 x 2^3 x 2^1

2^16 = 8 x 8 x 8 x 8 x 8 x 2

2^16 = 8^5 x 2

Note that we can get only 5 factors of 8 out of 2^16; there is a “leftover” 2 that cannot be used.

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What is the greatest value of x such that 8^x is a factor of 16! ? [#permalink]

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16 Dec 2016, 04:13
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Bunuel wrote:
What is the greatest value of x such that 8^x is a factor of 16! ?

A. 2
B. 3
C. 5
D. 6
E. 8

Since, we need to find the greatest value of x such that $$8^x$$ is a factor of $$16!$$, it is same as finding the highest power of 8 in $$16!$$

In any question where we need to find the highest power of a number which can divide a factorial (product of first n natural numbers) or in other words, is a factor of the given factorial, all we need to do is

1. Prime factorize the number whose highest power is to be found.
2. Find the highest power of each of the prime factors in the factorial
3. Calculate how many such numbers (whose highest power is to be found) can be created using the highest power of each of its prime factors.

Let's apply the above steps to solve this question and then we will look at a couple of questions where we can apply this learning.

We need to find the highest power of 8, so let's begin by doing

Step-1: Prime factorization of 8.
$$8 = 2 * 2 * 2 = 2^3$$

Step-2: Find the highest power of each of the prime factors in the factorial
Since 8 has only one prime factor 2, we need to find the highest power of 2 in 16!. Now how do we do so? There are 2 ways based on the same principle.

Method 1
Keep dividing the factorial successively by 2 and keep adding the quotient till you don't have anything left to divide. Remember, successive division means dividing the quotient obtained at each step by the same divisor by which we start the division.

So, let's do it quickly.
$$\frac {16}{2} = 8$$
$$\frac {8}{2} = 4$$
$$\frac {4}{2} = 2$$
$$\frac {2}{2} = 1$$
$$\frac {1}{2} = 0$$

As there is nothing left to divide, let's add the quotients to find the highest power of 2 in $$16!$$
Sum of quotients $$= 8+4+2+1+0 = 15$$

Method 2
Divide 16 by consecutive powers of 2, till you get 0 as a quotient and add all quotients. This method is based on the same principle as Method 1.

So, we have highest power of 2 in 16! = $$\frac {16}{2} + \frac {16}{2^2} +\frac {16}{2^3}+\frac {16}{2^4} = 8+4+2+1 = 15$$

So, we can conclude that the highest power of 2 in $$16!$$ is $$2^{15}$$

Step-3: Calculate how many such numbers can be created using the highest power of each of its prime factors
Let's try to figure how many 8's we can create using $$2^{15}$$.
Since, $$8 = 2^3$$, we can write $$2^{15}$$ as $$(2^{3})^5$$.
Or, in simple terms $$2^{15} = 8^5$$

Hence, the greatest value of x such that $$8^x$$ is a factor of 16! is 5. Hence, answer is choice C.

Let me post a couple of questions on similar lines where we can use the method discussed in this post to solve this type of questions very quickly.

Question 1: What is the greatest value of x such that $$15^x$$ completely divides 300! ?
A. 20
B. 54
C. 74
D. 148
E. 222

Question 2: What is the greatest value of "a" such that $$45^a$$ completely divides 300! ?
A. 8
B. 36
C. 74
D. 148
E. 222

Detailed solutions will be posted soon. Use the method highlighted in this post to solve the above questions. All the best.

To practise ten 700+ Level Number Properties Questions attempt the The E-GMAT Number Properties Knockout

Regards,
Piyush
e-GMAT
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Re: What is the greatest value of x such that 8^x is a factor of 16! ? [#permalink]

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16 Dec 2016, 09:11
Bunuel wrote:
What is the greatest value of x such that 8^x is a factor of 16! ?

A. 2
B. 3
C. 5
D. 6
E. 8

8^x can be simplified to 2^3x
largest power of 2 in 16!= 16/2 + 16/4 + 16/8 + 16/16= 8+4+2+1=15
8^x= 2^15
x= 15/3=5
C
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Re: What is the greatest value of x such that 8^x is a factor of 16! ? [#permalink]

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25 Dec 2016, 06:10
1
KUDOS
Expert's post
EgmatQuantExpert wrote:
Bunuel wrote:
What is the greatest value of x such that 8^x is a factor of 16! ?

A. 2
B. 3
C. 5
D. 6
E. 8

Since, we need to find the greatest value of x such that $$8^x$$ is a factor of $$16!$$, it is same as finding the highest power of 8 in $$16!$$

In any question where we need to find the highest power of a number which can divide a factorial (product of first n natural numbers) or in other words, is a factor of the given factorial, all we need to do is

1. Prime factorize the number whose highest power is to be found.
2. Find the highest power of each of the prime factors in the factorial
3. Calculate how many such numbers (whose highest power is to be found) can be created using the highest power of each of its prime factors.

Let's apply the above steps to solve this question and then we will look at a couple of questions where we can apply this learning.

We need to find the highest power of 8, so let's begin by doing

Step-1: Prime factorization of 8.
$$8 = 2 * 2 * 2 = 2^3$$

Step-2: Find the highest power of each of the prime factors in the factorial
Since 8 has only one prime factor 2, we need to find the highest power of 2 in 16!. Now how do we do so? There are 2 ways based on the same principle.

Method 1
Keep dividing the factorial successively by 2 and keep adding the quotient till you don't have anything left to divide. Remember, successive division means dividing the quotient obtained at each step by the same divisor by which we start the division.

So, let's do it quickly.
$$\frac {16}{2} = 8$$
$$\frac {8}{2} = 4$$
$$\frac {4}{2} = 2$$
$$\frac {2}{2} = 1$$
$$\frac {1}{2} = 0$$

As there is nothing left to divide, let's add the quotients to find the highest power of 2 in $$16!$$
Sum of quotients $$= 8+4+2+1+0 = 15$$

Method 2
Divide 16 by consecutive powers of 2, till you get 0 as a quotient and add all quotients. This method is based on the same principle as Method 1.

So, we have highest power of 2 in 16! = $$\frac {16}{2} + \frac {16}{2^2} +\frac {16}{2^3}+\frac {16}{2^4} = 8+4+2+1 = 15$$

So, we can conclude that the highest power of 2 in $$16!$$ is $$2^{15}$$

Step-3: Calculate how many such numbers can be created using the highest power of each of its prime factors
Let's try to figure how many 8's we can create using $$2^{15}$$.
Since, $$8 = 2^3$$, we can write $$2^{15}$$ as $$(2^{3})^5$$.
Or, in simple terms $$2^{15} = 8^5$$

Hence, the greatest value of x such that $$8^x$$ is a factor of 16! is 5. Hence, answer is choice C.

Let me post a couple of questions on similar lines where we can use the method discussed in this post to solve this type of questions very quickly.

Question 1: What is the greatest value of x such that $$15^x$$ completely divides 300! ?
A. 20
B. 54
C. 74
D. 148
E. 222

Question 2: What is the greatest value of "a" such that $$45^a$$ completely divides 300! ?
A. 8
B. 36
C. 74
D. 148
E. 222

Detailed solutions will be posted soon. Use the method highlighted in this post to solve the above questions. All the best.

To practise ten 700+ Level Number Properties Questions attempt the The E-GMAT Number Properties Knockout

Regards,
Piyush
e-GMAT

Alright so let's look at the detailed solution of the first question and the answer of the second question. Once you go through the above post and this solution, you should have a strong understanding of this approach. Let's use the simple 3-step approach to solve questions similar to the above question.

Step 1. Prime factorize the number whose highest power is to be found.
Step 2. Find the highest power of each of the prime factors in the factorial
Step 3. Calculate how many such numbers (whose highest power is to be found) can be created using the highest power of each of its prime factors.

Let's apply the above steps to solve this question and then we will look at a couple of questions where we can apply this learning.

We need to find the highest power of 15, so let's begin by doing

Step-1: Prime factorization of 15.
$$15 = 3*5 = 3^1*5^1$$

Step-2: Find the highest power of each of the prime factors in the factorial
Since 15 has two prime factors 3 and 5, we need to find the highest power of 3 and 5 separately in 300!. Now how do we do so? There are 2 ways based on the same principle.

Note: The question can be solved by finding the highest power of 5 alone. However, we don't advocate using this shortcut until you're 100% confident about these questions. The reason is there are certain complicated factors that you need to keep in mind to use this shortcut without making errors and the advantage gained is not big enough to take that risk. Hence, let's find the highest powers of both the prime factors 3 and 5.

Method 1
Keep dividing the factorial successively by 3 and keep adding the quotient till you don't have anything left to divide. Remember, successive division means dividing the quotient obtained at each step by the same divisor by which we start the division.

So, let's do it quickly.
$$\frac {300}{3} = 100$$
$$\frac {100}{3} = 33$$
$$\frac {33}{3} = 11$$
$$\frac {11}{3} = 3$$
$$\frac {3}{3} = 1$$
$$\frac {1}{3} = 0$$

As there is nothing left to divide, let's add the quotients to find the highest power of 3 in $$300!$$
Sum of quotients $$= 100+33+11+3+1+0 = 148$$

Similarly, let's find the highest power of 5 in 300! by using the same method.

$$\frac {300}{5} = 60$$
$$\frac {60}{5} = 12$$
$$\frac {12}{5} = 2$$
$$\frac {2}{5} = 0$$

As there is nothing left to divide, let's add the quotients to find the highest power of 5 in $$300!$$
Sum of quotients $$= 60+12+2+0 = 74$$

Method 2
Divide 300 by consecutive powers of 3, till you get 0 as a quotient and add all quotients. This method is based on the same principle as Method 1.

So, we have highest power of 3 in 300! = $$\frac {300}{3} + \frac {300}{3^2} +\frac {300}{3^3}+\frac {300}{3^4}+\frac {300}{3^5} = 100+33+11+3+1 = 148$$

So, we can conclude that the highest power of 3 in $$300!$$ is $$3^{148}$$

Following a similar process to find the highest power of 5 in 300!, we get that the highest power is $$5^{74}$$

Step-3: Calculate how many such numbers can be created using the highest power of each of its prime factors
Let's try to figure how many 15's we can create using $$3^{148}$$ and $$5^{74}$$.
Since, $$15 = 3^1*5^1$$, we can write $$300!$$ as $$300!=3^{148}*5^{74}*k = (3*5)^{74} * 3^{74}*k =15^{74}*3^{74}*k$$, where k is a positive integer
Or, in simple terms the highest power of 15 in 300 is $$15^{74}$$.

If use the same approach to solve Q2, you will understand that $$45 = 3^2*5^1$$.

The highest power of 3 in 300! is $$3^{148}$$ and that of 5 is $$5^{74}$$. Therefore, the highest power of 45 or $$3^2*5^1$$ in 300! is 74. Hence, the answer is choice C.

Regards,
Piyush
e-GMAT
_________________

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Re: What is the greatest value of x such that 8^x is a factor of 16! ? [#permalink]

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29 Mar 2017, 10:12
The key to this question is to first visualize what is going on
we have 16! = 16*15*14*....*3*2*1.

We need to find that max value of exponent x such that 8^x divides 16!.
let us see if we multiply 2 with 4 alone that will have one multiple of 8. We got first group of three 2s. We need to see how many groups can we construct of three 2s.
If we start summing the exponent of 2 in every even number we will get 15. In 2 we have just 1, in 4 we got 2, in 8 we got 3 and so on. Now since 15/3 = 5. Since we are looking for batch of three 2s. 5 should be the answer.
Re: What is the greatest value of x such that 8^x is a factor of 16! ?   [#permalink] 29 Mar 2017, 10:12
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