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# In the expression 13!/x^3, for which of the following values of x will

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In the expression 13!/x^3, for which of the following values of x will  [#permalink]

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23 Jun 2017, 02:34
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In the expression $$\frac{13!}{x^3}$$, for which of the following values of x will the expression will NOT be an integer?

A. 3
B. 4
C. 5
D. 6
E. 8

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In the expression 13!/x^3, for which of the following values of x will  [#permalink]

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23 Jun 2017, 02:51
$$13! = 13*12*11*10*9*8*7*6*5*4*3*2*1 = 13*12*11*7*5^2*3^4*2^8$$

Of the answer options for values of x, at x=5
the expression $$\frac{13!}{x^3}$$ will not be an integer.

Hence Option C is the correct answer.
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In the expression 13!/x^3, for which of the following values of x will  [#permalink]

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23 Jun 2017, 07:28
Bunuel wrote:
In the expression $$\frac{13!}{x^3}$$, for which of the following values of x will the expression will NOT be an integer?

A. 3
B. 4
C. 5
D. 6
E. 8

Method I - Raise each answer choice to the power of 3 because we're looking for $$x^3.$$. Prime factorize each answer choice if not prime.

Then find the number of powers of the prime(s) in 13! The answer will be an integer if and only if there are enough powers of the prime factor(s) in the factorial. Using formula:

$$\frac{13}{n} + \frac{13}{n^2} + \frac{13}{n^3}$$ . . . only to the point where n raised to whatever power is ≤ 13.**

A. $$3^3$$. We need three powers of 3.

$$\frac{13}{3^1}= 4$$

$$\frac{13}{3^2} = 1$$

13 cannot be divided by $$3^3$$ to get an integer, so stop there.

There are 4 + 1 = 5 powers of 3 in 13!. We only need 3. $$\frac{13!}{3^3}$$ will be an integer.

B. $$4^3 = (2^2)^3 = 2^6$$. We need six powers of 2.

$$\frac{13}{2}$$ = 6

$$\frac{13}{2^2}$$ = 3

$$\frac{13}{2^3 = 1}$$
, so

There are 6 + 3 + 1 = 10 powers of 2 in 13! We need six powers of 2 to make $$4^3$$. We have ten. $$\frac{13!}{4^3}$$ will be an integer.

C. $$5^3$$

$$\frac{13}{5} = 2$$

13 cannot be divided by $$5^2$$ to get an integer.

We only have two powers of 5 and we need three. $$\frac{13!}{5^3}$$ will not be an integer.

D. $$6^3 = 2^3*3^3$$. Check to see that there are enough powers of both 2 and 3.

From answer B, there are ten powers of 2, and we need only three.

From answer A, there are five powers of 3, also enough. $$\frac{13!}{6^3}$$ will be an integer.

E.$$8^3 = (2^3)^3$$, or $$2^9$$

From Answer B: there are ten powers of 2, and we need only nine. $$\frac{13!}{8^3}$$will be an integer.

Method II - Look at the factors of 13! and see whether or not for each answer choice there are three powers of the number in 13!

13! = 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.

A.$$3^3$$. Number of powers of 3 in 13!
3 has one three
6 has one three
9 has two threes
12 has one three
Total of five powers of 3. We only need three to get even division and thereby an integer. $$\frac{13!}{3^3}$$ will be an integer.

B. $$4^3$$. Either prime factorize 4, or notice:
4 has one four
8 has two fours
12 has one four. We have the necessary three powers of 4. $$\frac{13!}{4^3}$$ will be an integer.

C. $$5^3$$
5 has one five
10 has one five
There are no other powers of 5. We have two, we need three. $$fraction]13!/5^3$$[/fraction] will not be an integer.

D. $$6^3$$ = $$2^3*3^3$$. There aren't enough obvious powers of 6 in 13! There is one in 6, and one in 12. We need three powers of 6.

But by prime factorizing $$6^3$$ to $$2^33^3$$

From answer A, there are five powers of 3.
We can get one power of 2 from 2, and two powers of 2 from 4, total three as needed.

There are enough powers of 2 and 3 that $$\frac{13!}{6^3}$$ will be an integer.

E.$$8^3 = (2^3)^3 = 2^9$$
2 has one two
4 has two twos
6 has one two
8 has three twos
10 has one two
12 has two twos. That's 10 total. We need 9.$$\frac{13!}{8^3}$$ will be an integer.

(C)

** See Bunuel 's excellent Math Number Theory, here https://gmatclub.com/forum/math-number-theory-88376.html, and scroll down to Factorials, Finding the number of powers of a prime number $$p$$, in the $$n!$$
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Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

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23 Jun 2017, 09:37
Bunuel wrote:
In the expression $$\frac{13!}{x^3}$$, for which of the following values of x will the expression will NOT be an integer?

A. 3
B. 4
C. 5
D. 6
E. 8

Since we are dividing by 3rd power of x.
Hence there should be atleast 3 powers of x in the numerator:

x = 3 -> 13! has more than 3 powers of 3. Integer
x = 4 -> 13! has more than 3 powers of 4. Integer
x = 5 -> 13! has 2 powers of 5. Not an integer.

Correct Option C
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Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

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23 Jun 2017, 09:42
13! = 13*12*11*10...*5*4..*2*1

The number of 5's in 13! are only '2': one in '10' and the other in '5'. So 13! will NOT be divisible by 5^3.

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Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

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26 Jun 2017, 16:57
Bunuel wrote:
In the expression $$\frac{13!}{x^3}$$, for which of the following values of x will the expression will NOT be an integer?

A. 3
B. 4
C. 5
D. 6
E. 8

Let’s prime factorize 13!.

13! = 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2

13! = 13 x (2^2 x 3^1) x 11 x (5 x 2) x 3^2 x 2^3 x 7 x (2 x 3) x 5 x 2^2 x 3 x 2

13! = 13 x 11 x 7 x 5^2 x 3^5 x 2^10

Let’s now analyze each answer choice:

A) x = 3

Thus, x^3 = 3^3.

Since there are five 3s in 13!, x can be 3.

B) x = 4 = 2^2

Thus, x^3 = 2^6.

Since there are ten 2s in 13!, x can be 4.

C) x = 5

Thus, x^3 = 5^3.

Since there are only two 5s in 13!, x cannot be 5.

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Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

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27 Jun 2017, 09:49
Imo C
The power of 3 in 13!=5
The power of 4 in 13!=3
The power of 5 in 13!=2
The power of 6 in 13!=2
The power of 8 in 13!=1
The power of 2 in 13!=10
So 3! will be divisible by 27 ,64,512,216 but will not be divisible by 125 so C is the answer.
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Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

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27 Jun 2017, 11:36
1
In the expression $$\frac{13!}{x^3}$$, for which of the following values of x will the expression will NOT be an integer?

In order to find what values of x will NOT be an integer, lets first find out how are we going to get the integer.

So, in order to get an integer we want denominator to completely divide the numerator, in other words can we change the denominator to 1

This is only possible if there are sufficient factors which can convert all the three units of denominator numbers to 1

Lets evaluate the answer choices, if they have at least three factors in order to completely divide the numerator.

A. 3 =====> Factors in numerator - 3, 6 & 9 ===> Gives Integer
B. 4 =====> Factors in numerator - 4, 8 & 12 ===> Gives Integer
C. 5 =====> Factors in numerator = 5 & 10 ===> Does not Give Integer as Third Unit of denominator cannot completely divide the numerator.
D. 6 =====> Factors in numerator = 2*3, 6, 12 ===> Gives Integer
E. 8 =====> Factors in numerator = 2*4, 8,4 (From 12), 2 (From 10), 2 (From 6) {4*2*2 = 8} ===> Gives Integer

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Re: In the expression 13!/x^3, for which of the following values of x will   [#permalink] 19 Jan 2019, 03:17
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