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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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19 Dec 2009, 15:26
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95% (hard)

Question Stats:

29% (01:41) correct 71% (01:51) wrong based on 450 sessions

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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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21 Dec 2009, 04:05
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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 $$z=1$$.

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Re: PR1012 - Factorials 4 [#permalink]

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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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22 Dec 2009, 09:12
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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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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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29 Aug 2014, 03:17
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Re: If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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

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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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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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23 Jul 2017, 20:37
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Re: If x, y, and z are positive integers, and (x!+x)/z=y, then [#permalink]

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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 $$z=1$$.

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

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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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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