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For integers x and y, 1 < x < y. Is xy a factor of 11!?

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For integers x and y, 1 < x < y. Is xy a factor of 11!? [#permalink]

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New post 29 Jul 2015, 15:52
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Question Stats:

45% (01:17) correct 55% (01:01) wrong based on 219 sessions

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Re: For integers x and y, 1 < x < y. Is xy a factor of 11!? [#permalink]

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Bunuel wrote:
For integers x and y, 1 < x < y. Is x^y a factor of 11!?

(1) x > 3
(2) x is prime


Kudos for a correct solution.


Given: y>x>1

Is x^y a factor of 11! => Can x^y a divisible of 11!

1. x >3

Since y>x,

Assume x=4 and y = 5, 4^5=>2^10

But 11! has only 2^8. Hence 4^5 is not a factor of 11!

if we choose any number bigger than this, still x^y will not be a factor of 11!

Hence Sufficient.

2. x is prime => x could be 2, 3, 5, 7, 11

2^3 is a factor of 11!
But 5^6 is not.

Hence not sufficient.

Hence Option A
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Re: For integers x and y, 1 < x < y. Is xy a factor of 11!? [#permalink]

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New post 30 Jul 2015, 02:07
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Bunuel wrote:
For integers x and y, 1 < x < y. Is x^y a factor of 11!?

(1) x > 3
(2) x is prime


Kudos for a correct solution.


IMO : A

Question Stem: 1 < x < y.
11! = \(2^8 * 3^4 * 5^2 * 7 * 11\)

Statement 1: x>3
for x>3 the power of x i.e y will be equal to or less than x
Eg: if x = 4, then 2^8 can be written in terms of 4 =4^4. Where x = y
for rest of the values x>y(which contradicts the condition given in the question stem)
Thus, no value of x for x>3 \(x^y\) is factor of 11!
Hence Suff

Statement 2: x is prime
satisfies when x = 2,3
doesn't satisfy when x=5,7,11
Hence not suff
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Re: For integers x and y, 1 < x < y. Is xy a factor of 11!? [#permalink]

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Bunuel wrote:
For integers x and y, 1 < x < y. Is x^y a factor of 11!?

(1) x > 3
(2) x is prime


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

For this problem, it's helpful to factor out 11! to see what factors it will contain. 11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1.

With that, you should see relatively quickly that certain values of x cannot be taken to an exponent greater than 1 and still remain a factor of 11!. 11 and 7, for example, are prime numbers that only appear once. So because the base of the exponent (x) has to be taken to an exponent greater than itself (y), those numbers are certainly eliminated.

So with that in mind, you'll need a relatively low value of x so that you can maintain a relatively low exponent, too. 2 would work (in 11!, 10 (which is 5 * 2), 8 (which is 2 * 2 * 2), 6 (which is 2 * 3), 4 (which is 2 * 2), and 2 all give you factors of 2, so you can take 2 all the way to the 8th power and it will still work). And 3 would work (3, 6, and 9 - which is 3 * 3 and therefore gives you two 3s - are part of 11!, so you can take 3 to the 4th power, too). But that's it.

So when statement 1 tells you that the lowest possible value of x is 4, then you should consider 4 to the 5th. 4 can be expressed as 2 * 2, so that leaves you with 2 to the 10th power - and since there are only eight 2s contained within 11!, you cannot have x = 4 and y = 5 as options. So statement 1 is sufficient.

Statement 2 is not sufficient, as it allows for x = 2 and y = "anything between 2 and 9" (all of which would give you "yes") but also allows for a prime number of x like 11 (which would mean "no").
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Re: For integers x and y, 1 < x < y. Is xy a factor of 11!? [#permalink]

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New post 23 Dec 2017, 12:15
Bunuel wrote:
Bunuel wrote:
For integers x and y, 1 < x < y. Is x^y a factor of 11!?

(1) x > 3
(2) x is prime


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

For this problem, it's helpful to factor out 11! to see what factors it will contain. 11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1.

With that, you should see relatively quickly that certain values of x cannot be taken to an exponent greater than 1 and still remain a factor of 11!. 11 and 7, for example, are prime numbers that only appear once. So because the base of the exponent (x) has to be taken to an exponent greater than itself (y), those numbers are certainly eliminated.

So with that in mind, you'll need a relatively low value of x so that you can maintain a relatively low exponent, too. 2 would work (in 11!, 10 (which is 5 * 2), 8 (which is 2 * 2 * 2), 6 (which is 2 * 3), 4 (which is 2 * 2), and 2 all give you factors of 2, so you can take 2 all the way to the 8th power and it will still work). And 3 would work (3, 6, and 9 - which is 3 * 3 and therefore gives you two 3s - are part of 11!, so you can take 3 to the 4th power, too). But that's it.

So when statement 1 tells you that the lowest possible value of x is 4, then you should consider 4 to the 5th. 4 can be expressed as 2 * 2, so that leaves you with 2 to the 10th power - and since there are only eight 2s contained within 11!, you cannot have x = 4 and y = 5 as options. So statement 1 is sufficient.

Statement 2 is not sufficient, as it allows for x = 2 and y = "anything between 2 and 9" (all of which would give you "yes") but also allows for a prime number of x like 11 (which would mean "no").


2,3 & 5 are also primes, why aren't those also considered?
- I don't understand why the lowest value of x=4
Re: For integers x and y, 1 < x < y. Is xy a factor of 11!?   [#permalink] 23 Dec 2017, 12:15
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