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If x, y, and z are positive integers, and (x!+x)/z=y, then

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If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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If x, y, and z are positive integers, and \(\frac{x!+x}{z}=y\), then what is the value of z?

(1) x is a factor of y
(2) z < x

[Reveal] Spoiler:
With the first clue I get to the form \((x-1)!/c+1/c=z\), where c is a positive integer, but I don't know how to use the second clue.
[Reveal] Spoiler: OA

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Re: PR1012 - Factorials 4 [#permalink]

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Marco83 wrote:
I would like an explanation on how to solve this exercise without plugging numbers.

If x, y, and z are positive integers, and \((x!+x)/z=y\), then what is the value of z?

I) x is a factor of y

II) z<x


With the first clue I get to the form \((x-1)!/c+1/c=z\), where c is a positive integer, but I don't know how to use the second clue.


First note that for \(n>1\), \(n!+1\) has no factor \(k<=n\), but \(1\). As \(n!=2*3*...*n\), which means that every integer less than or equal to \(n\) is a factor of \(n!\), but if we add \(1\) to \(n!\), sum, \(n!+1\), obviously won't be evenly divisible by these factors.

For example:
\(n=3\), \(n!+1=7\), \(7\) has no factor \(<=3\), except \(1\).
\(n=4\), \(n!+1=25\), \(25\) has no factor \(<=4\), except \(1\).
\(n=5\), \(n!+1=121\), \(121\) has no factor \(<=5\), except \(1\).

Back to the question:

If \(x\), \(y\), and \(z\) are positive integers, and \(\frac{x!+x}{z}=y\), then what is the value of \(z\)?

(1) \(x\) is a factor of \(y\) --> \(xk=y\) --> \(x!+x=yz=xkz\) --> \((x-1)!+1=kz\) --> \(z=1\) OR \(z>x-1\). Not sufficient.

For example:
\(x=4\), \((x-1)!+1=7=kz\), \(z=1\) OR \(z=7>4-1=3\)
\(x=5\), \((x-1)!+1=25=kz\), \(z=1\) OR \(z=5>5-1=4\) OR \(z=25>5-1=4\)
\(x=6\), \((x-1)!+1=121=kz\), \(z=1\) OR \(z=11>5-1=4\) OR \(z=121>5-1=4\)

(2) \(z<x\). Clearly insufficient.

For example:
\(x=4\), \(x!+x=28=yz\), \(z=1\) OR \(z=2\) OR \(z=4\) OR \(z=14\) OR \(z=28\).

(1)+(2) \(z=1\) OR \(z>x-1\) AND \(z<x\) --> As \(z<x\) and \(z\) is an integer, \(z>x-1\) can not be true, hence \(z=1\).

Answer: C.
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Re: PR1012 - Factorials 4 [#permalink]

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New post 22 Dec 2009, 08:34
@bunuel i dont get why z > x-1 please can you explain?

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Re: PR1012 - Factorials 4 [#permalink]

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sagarsabnis wrote:
@bunuel i dont get why z > x-1 please can you explain?


We know that \(n!+1\) has no factor \(k\), such as \(k<=n\), but \(1\). Refer to the explanation in my previous post.

We have \((x-1)!+1=kz\), which means that \(z\) is a factor of \((x-1)!+1\), hence either \(z\) equals to \(1\) OR \(z>x-1\), as \(z\) can not be \(<=x-1\) (see above).

Hope it's clear.
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Re: PR1012 - Factorials 4 [#permalink]

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New post 22 Dec 2009, 11:39
now its clear...thanks

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Re: If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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Re: If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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New post 30 Sep 2015, 18:43
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Re: If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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New post 10 Nov 2015, 16:16
Hi Bunuel,

how do you get from here "x!+x=yz=xkz" to here "(x−1)!+1=kz" ?

Thanks so much!

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New post 10 Nov 2015, 23:28
angierch wrote:
Hi Bunuel,

how do you get from here "x!+x=yz=xkz" to here "(x−1)!+1=kz" ?

Thanks so much!


First of all: \(x! = (x-1)!*x\), the same way for example, \(6!=(6-1)!*6=5!*6\).

Thus \(x!+x=(x-1)!*x+x=x((x-1)!+1)\). So, we have that \(x((x-1)!+1)=xkz\). Reduce by x to get \((x-1)!+1)=kz\).

Hope it's clear.
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Re: If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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New post 11 Nov 2015, 09:10
It is clear. Thank you very much!

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Re: If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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New post 02 Aug 2017, 00:11
Bunuel wrote:
Marco83 wrote:
I would like an explanation on how to solve this exercise without plugging numbers.

If x, y, and z are positive integers, and \((x!+x)/z=y\), then what is the value of z?

I) x is a factor of y

II) z<x


With the first clue I get to the form \((x-1)!/c+1/c=z\), where c is a positive integer, but I don't know how to use the second clue.


First note that for \(n>1\), \(n!+1\) has no factor \(k<=n\), but \(1\). As \(n!=2*3*...*n\), which means that every integer less than or equal to \(n\) is a factor of \(n!\), but if we add \(1\) to \(n!\), sum, \(n!+1\), obviously won't be evenly divisible by these factors.

For example:
\(n=3\), \(n!+1=7\), \(7\) has no factor \(<=3\), except \(1\).
\(n=4\), \(n!+1=25\), \(25\) has no factor \(<=4\), except \(1\).
\(n=5\), \(n!+1=121\), \(121\) has no factor \(<=5\), except \(1\).

Back to the question:

If \(x\), \(y\), and \(z\) are positive integers, and \(\frac{x!+x}{z}=y\), then what is the value of \(z\)?

(1) \(x\) is a factor of \(y\) --> \(xk=y\) --> \(x!+x=yz=xkz\) --> \((x-1)!+1=kz\) --> \(z=1\) OR \(z>x-1\). Not sufficient.

For example:
\(x=4\), \((x-1)!+1=7=kz\), \(z=1\) OR \(z=7>4-1=3\)
\(x=5\), \((x-1)!+1=25=kz\), \(z=1\) OR \(z=5>5-1=4\) OR \(z=25>5-1=4\)
\(x=6\), \((x-1)!+1=121=kz\), \(z=1\) OR \(z=11>5-1=4\) OR \(z=121>5-1=4\)

(2) \(z<x\). Clearly insufficient.

For example:
\(x=4\), \(x!+x=28=yz\), \(z=1\) OR \(z=2\) OR \(z=4\) OR \(z=14\) OR \(z=28\).

(1)+(2) \(z=1\) OR \(z>x-1\) AND \(z<x\) --> As \(z<x\) and \(z\) is an integer, \(z>x-1\) can not be true, hence \(z=1\).

Answer: C.


This doesn't feel very GMAT-like. Isn't this too tricky to be solvedd under 2 minutes? :oops:

chetan2u , can such questions be considered in a GMAT exam?
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If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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New post 09 Aug 2017, 05:29
x!+x/z=y
1) x is a factor of y.
let x=3! we have 3!+3/z=y or 9/z=y
z can be 1, 3 or 9 ie 3 values.
let x=2! we have 2!+3/z=y or 5/z=y
This is not valid since x is not a factor of 5.
not sufficient
2) x<y not sufficient as x and y can have multiple values and subsequently z will have multiple values hence not sufficient

combine 1&2. x is a factor of y ie 3 is a factor of 9 and x has to be less than y ie 9.
Now if z=3 or 9 than y=3 or 1 which is not valid as per statement 2. Hence z=1.
combined sufficient.

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If x, y, and z are positive integers, and (x!+x)/z=y, then   [#permalink] 09 Aug 2017, 05:29
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