Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 21 Jul 2019, 04:05 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x, y, and z are positive integers, and (x!+x)/z=y, then

Author Message
TAGS:

### Hide Tags

Intern  Joined: 08 Nov 2009
Posts: 43
If x, y, and z are positive integers, and (x!+x)/z=y, then  [#permalink]

### Show Tags

4
41 00:00

Difficulty:   95% (hard)

Question Stats: 28% (02:36) correct 72% (02:21) wrong based on 528 sessions

### HideShow timer Statistics 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

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 56306
Re: PR1012 - Factorials 4  [#permalink]

### Show Tags

7
9
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 $$z=1$$.

_________________
##### General Discussion
Manager  Joined: 22 Jul 2009
Posts: 132
Location: Manchester UK
Re: PR1012 - Factorials 4  [#permalink]

### Show Tags

@bunuel i dont get why z > x-1 please can you explain?
Math Expert V
Joined: 02 Sep 2009
Posts: 56306
Re: PR1012 - Factorials 4  [#permalink]

### Show Tags

1
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.
_________________
Manager  Joined: 22 Jul 2009
Posts: 132
Location: Manchester UK
Re: PR1012 - Factorials 4  [#permalink]

### Show Tags

now its clear...thanks
Intern  Joined: 11 Jun 2015
Posts: 9
Re: If x, y, and z are positive integers, and (x!+x)/z=y, then  [#permalink]

### Show Tags

Hi Bunuel,

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

Thanks so much!
Math Expert V
Joined: 02 Sep 2009
Posts: 56306
Re: If x, y, and z are positive integers, and (x!+x)/z=y, then  [#permalink]

### Show Tags

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.
_________________
Intern  Joined: 11 Jun 2015
Posts: 9
Re: If x, y, and z are positive integers, and (x!+x)/z=y, then  [#permalink]

### Show Tags

It is clear. Thank you very much!
Manager  B
Joined: 04 May 2014
Posts: 158
Location: India
WE: Sales (Mutual Funds and Brokerage)
If x, y, and z are positive integers, and (x!+x)/z=y, then  [#permalink]

### Show Tags

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.
Non-Human User Joined: 09 Sep 2013
Posts: 11720
Re: If x, y, and z are positive integers, and (x!+x)/z=y, then  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: If x, y, and z are positive integers, and (x!+x)/z=y, then   [#permalink] 27 Dec 2018, 10:37
Display posts from previous: Sort by

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