GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Oct 2019, 18:25 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.  How many positive integral values of N are there such that

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1022
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
How many positive integral values of N are there such that  [#permalink]

Show Tags

1
9 00:00

Difficulty:   95% (hard)

Question Stats: 33% (02:19) correct 67% (02:36) wrong based on 118 sessions

HideShow timer Statistics

How many positive integral values of N are there such that $$\frac{N+76}{N+4}$$ is an integer?

a) 5
b) 7
c) 8
d) 9
e) 10

_________________
Press +1 Kudos If my post helps!
Manager  P
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 103
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

4
6
$$\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}$$

$$1$$ is an integer, so let's see for how many values of N, $$\frac{72}{N+4}$$ is an integer.

$$72 = 2^3*3^2$$

Thus the number of factors (numbers that can divide $$72$$) is equal to $$(3+1)*(2+1)=12$$

So $$72$$ has $$12$$ factors, but $$N+4$$ is greater than $$4$$ so we have to remove the factors that are less or equal to $$4$$.

So we have to remove, $$1$$, $$2$$, $$3$$ and $$4$$ (those are the factors of $$72$$ that are less or equal than $$4$$)

So $$12-4=8$$

Don't hesitate to ask if something is not clear.
_________________
General Discussion
Manager  B
Joined: 28 Jun 2018
Posts: 73
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

AnisMURR How did you get "Thus the number of factors (numbers that can divide 7272) is equal to (3+1)∗(2+1)=12(3+1)∗(2+1)=12"
Manager  P
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 103
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

1
hibobotamuss

$$72=2^3*3^2$$

You take all the powers of the prime factors you add 1, and then you multiply them. It is a formula.

lets say $$X=2^3*3^2*5^5*7^4$$

The number of factors is $$4*3*6*5$$

Don't hesitate if it is not clear, you can also visit my youtube channel for more exercices.
_________________
Manager  B
Joined: 29 May 2017
Posts: 125
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

AnisMURR wrote:
$$\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}$$

$$1$$ is an integer, so let's see for how many values of N, $$\frac{72}{N+4}$$ is an integer.

$$72 = 2^3*3^2$$

Thus the number of factors (numbers that can divide $$72$$) is equal to $$(3+1)*(2+1)=12$$

So $$72$$ has $$12$$ factors, but $$N+4$$ is greater than $$4$$ so we have to remove the factors that are less or equal to $$4$$.

So we have to remove, $$1$$, $$2$$, $$3$$ and $$4$$ (those are the factors of $$72$$ that are less or equal than $$4$$)

So $$12-4=8$$

Don't hesitate to ask if something is not clear.

when you say N+4 is greater than 4...you are excluding negative values of N.

Can you explain why I cant have N<0?

regards
Manager  S
Joined: 08 Sep 2008
Posts: 140
Location: India
Concentration: Operations, General Management
Schools: ISB '20
GPA: 3.8
WE: Operations (Transportation)
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

Mansoor50 wrote:
AnisMURR wrote:
$$\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}$$

$$1$$ is an integer, so let's see for how many values of N, $$\frac{72}{N+4}$$ is an integer.

$$72 = 2^3*3^2$$

Thus the number of factors (numbers that can divide $$72$$) is equal to $$(3+1)*(2+1)=12$$

So $$72$$ has $$12$$ factors, but $$N+4$$ is greater than $$4$$ so we have to remove the factors that are less or equal to $$4$$.

So we have to remove, $$1$$, $$2$$, $$3$$ and $$4$$ (those are the factors of $$72$$ that are less or equal than $$4$$)

So $$12-4=8$$

Don't hesitate to ask if something is not clear.

when you say N+4 is greater than 4...you are excluding negative values of N.

Can you explain why I cant have N<0?

regards

As per question Only positive value of N to be considered.

Sent from my iPad using GMAT Club Forum mobile app
_________________
Regards;
Vishal

+==========+
Kudos if it deserve....one...
Manager  B
Joined: 29 May 2017
Posts: 125
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

vishalkazone wrote:
Mansoor50 wrote:
AnisMURR wrote:
$$\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}$$

$$1$$ is an integer, so let's see for how many values of N, $$\frac{72}{N+4}$$ is an integer.

$$72 = 2^3*3^2$$

Thus the number of factors (numbers that can divide $$72$$) is equal to $$(3+1)*(2+1)=12$$

So $$72$$ has $$12$$ factors, but $$N+4$$ is greater than $$4$$ so we have to remove the factors that are less or equal to $$4$$.

So we have to remove, $$1$$, $$2$$, $$3$$ and $$4$$ (those are the factors of $$72$$ that are less or equal than $$4$$)

So $$12-4=8$$

Don't hesitate to ask if something is not clear.

when you say N+4 is greater than 4...you are excluding negative values of N.

Can you explain why I cant have N<0?

regards

As per question Only positive value of N to be considered.

Sent from my iPad using GMAT Club Forum mobile app

Thanks!!!!

note to self: Mansoor...read the question CAREFULLY!!!! Manager  P
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 103
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

vishalkazone
Mansoor50

"How many positive integral values of N " N should be positive

Don't hesitate if it is not clear, you can also visit my youtube channel for more exercices.
_________________
Manager  G
Joined: 21 Jun 2017
Posts: 230
Concentration: Finance, Economics
WE: Corporate Finance (Commercial Banking)
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

chetan2u Any other approach ?
_________________
Even if it takes me 30 attempts, I am determined enough to score 740+ in my 31st attempt. This is it, this is what I have been waiting for, now is the time to get up and fight, for my life is 100% my responsibility.

Dil ye Ziddi hai !!!

GMAT 1 - 620 .... Disappointed for 6 months. Im back Im back. Bhai dera tera COMEBACK !!!
Math Expert V
Joined: 02 Aug 2009
Posts: 7959
Re: How many positive integral values of N are there such that  [#permalink]

Show Tags

2
1
Afc0892 wrote:
How many positive integral values of N are there such that $$\frac{N+76}{N+4}$$ is an integer?

a) 5
b) 7
c) 8
d) 9
e) 10

Ok ShankSouljaBoi,

Two ways,
(I) I will go with this, as also discussed above and would recommend to anyone who is aware of the method..
$$\frac{N+76}{N+4}= \frac{N+4}{N+4} + \frac{72}{N+4}$$
Now 72 should be a multiple of N+4....

So, our next step should be to find the factors of 72.
$$72=2^3*3^2$$.... Number of factors = (3+1)(2+1)=4*3=12
so N+4 can be equal to any of these 12 factors.

As, N is positive so N+4 will have the least value of 1+4 or 5, when N is 1.
Keeping in mind the constraint, all factors below 5 are not possible.
WHAT are the factors below 5 - they are 1, 2, 3, 4... so 4 of them

(II) If you are stuck..
any of such questions should be made equal to a variable.. so $$\frac{N+76}{N+4}=a.......N+76=aN+4a..........N(a-1)=76-4a.......N = \frac{76-4a}{a-1}$$
since N is integer, [$$fraction]76-4a/a-1[/fraction]$$ must be an integer.
so FIRST condition - \frac{numerator}{geq {denominator}} thus $$76-4a/geq{a-1}........a<\frac{77}{5}......a<15.4$$, thus greatest value of a is 15
so you have to just substitute a from 1 to 15 and see which all values give us an integer.

for example a = 1 not possible...
a=2, 76-4a=76-8=68 and a-1=2-1=1... 68 divisible by 1 ..yes
a=3, 76-4a=76-12=64 and a-1=3-1=2... 64 divisible by 2 ..yes
a=4, 76-4a=76-16=60 and a-1=4-1=3... 60 divisible by 3 ..yes
a=5, 76-4a=76-20=56 and a-1=5-1=4... 56 divisible by 4 ..yes
a=6, 76-4a=76-24=52 and a-1=6-1=5... 52 is not divisible by 3 ..
a=7, 76-4a=76-28=48 and a-1=7-1=6... 48 divisible by 3 ..yes
and so on..
Although it will not take too much time but let us see if we can shorten it..
when a is ODD, both are even so 3, 5, 7, 9, 13 will give you an integer.
they leave the same remainder when divided by 3 ... 76-4a or 75+1-3a-a will leave a remainder of 1-a as 75-3a will be divisible by 3 and the denominator is also a-1..
so when a is 4, 4+3, 7+3, 10+3 so 4,7,10,13
so 2, 3, 4, 5, 7, 9, 10, 13.. thus 8 of them ..
you can try for values of 6 or 11, but they both will not be divisible by 3 together.
_________________ Re: How many positive integral values of N are there such that   [#permalink] 11 Nov 2018, 09:51
Display posts from previous: Sort by

How many positive integral values of N are there such that

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  