GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 07 Aug 2020, 10:51 ### 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 an integer n is to be selected at random from 1 to 100, inclusive,

Author Message
TAGS:

### Hide Tags

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

2
Keats wrote:
If I see a sequence of 4 numbers, say for example, 1,2,3, and 4, we have two numbers that satisfy the case here - 3 and 4. So 2/4 makes sense.

DensetsuNo wrote:
eg. If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

Sequence of 8 numbers: 8
Numbers we have: 3

In this case, lets take a sequence of 8 - 1,2,3,4,5,6,7, and 8 - we have four numbers that satisfy the case - 2,4,6, and 8. So 4/8=1/2 should be the case. Why do you say it is 3/8?

There are two ways in which n(n + 1)(n + 2) can be a multiple of 8.

Case 1: n and (n+2) are even. Two consecutive even numbers will have exactly one multiple of 4. So out of the two - n and (n+2) - exactly one of them will be a multiple of 4 and the other a multiple of 2. So their product will be a multiple of 8.
Out of 96 numbers, exactly half will have n as even. (n+1) will be odd here but will be immaterial.
So this will be 1/2.

Case 2: (n+1) is even and a multiple of 8.
n can take values from 1 to 96. So (n+1) can take values from 2 to 97.
How many multiples of 8 lie between 2 and 97, inclusive? 8, 16, 24, 32, ... 96
8 * 1 = 8
8 * 2 = 16
...
8 * 12 = 96

So in another 12/96 = 1/8 cases, n(n+1)(n+2) will be a multiple of 8.

Total, in 1/2 + 1/8 = 5/8 of the cases, n(n+1)(n+2) will be a multiple of 8.
_________________
Karishma
Veritas Prep GMAT Instructor

Director  G
Joined: 28 Nov 2014
Posts: 805
Concentration: Strategy
Schools: Fisher '19 (M\$)
GPA: 3.71
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

Thanks VeritasPrepKarishma I understand the way you have done it. Even I did it the same way. However, I am perplexed how DensetsuNo is arriving onto the correct answer in just 10 seconds. I also want to learn this quick thing!
Intern  Joined: 21 Jun 2014
Posts: 27
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

mitko20m wrote:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

To find total number of cases where n(n+1) will be divisible by 4 =

Where n is divisible by 4 = 4,8,12,...100 = (100 - 4) / 4 + 1= 25. There 25 such numbers
Now cases where n+1 is divisible by 4, which is basically all numbers divisible by 3 = 3,6,9....99 = (99-3)/3 + 1 = 25

Total Cases = 50
Probability = 50/100

Took time for me to realize that I am solving it for n and n+1, but not just n

Regards
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

2
1
Keats wrote:
Thanks VeritasPrepKarishma I understand the way you have done it. Even I did it the same way. However, I am perplexed how DensetsuNo is arriving onto the correct answer in just 10 seconds. I also want to learn this quick thing!

In the sequence of 8 numbers where n = 1 to 8, there are 5 values for which n*(n+1)*(n+2) is divisible by 8.

1*2*3
2*3*4 - divisible by 8
3*4*5
4*5*6 - divisible by 8
5*6*7
6*7*8 - divisible by 8
7*8*9 - divisible by 8
8*9*10 - divisible by 8

For values of n from 9 to 16, we will have the same pattern and so on...

Probablity = 5/8
_________________
Karishma
Veritas Prep GMAT Instructor

Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11419
Location: United States (CA)
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

mitko20m wrote:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

We are given that an integer n is to be selected at random from 1 to 100, inclusive, and we must determine the probability that n(n+1) will be divisible by 4.

Since probability = favorable outcomes/total outcomes, and we know that the total number of outcomes is 100, since there are 100 integers from 1 to 100, inclusive, we need to determine the number of values of n such that n(n+1) is divisible by 4.

First, we can determine the number of values of n that are divisible by 4, that is, the number of multiples of 4 that are between 1 and 100, inclusive. To calculate this, we can use the formula:

(Largest multiple of 4 – smallest multiple of 4)/4 + 1

(100 - 4)/4 + 1

96/4 + 1

24 + 1 = 25

Thus, there are 25 multiples of 4 between 1 and 100 inclusive. That is, the value of n can be any one of these 25 multiple of 4 so that n(n + 1) will be divisible by 4.

Similarly, if n + 1 is a multiple of 4, n(n + 1) will be also be divisible by 4. Since we know that there are 25 values of n that are multiples of 4, there must be another 25 values of n such that n + 1 is a multiple of 4. Let’s expand on this idea:

When n = 3, n + 1 = 4, and thus n(n+1) is a multiple of 4…..

When n = 23, n + 1 = 24, and thus n(n+1) is a multiple of 4…..

When n = 99, n + 1 = 100, and thus n(n+1) is a multiple of 4.

So we can see that there are 25 values of n that are multiples of 4 and 25 more values of n for n + 1 that are multiples of 4. Thus, the probability of selecting a value of n so that n(n+1) is a multiple of 4 is:

50/100 = 1/2

_________________

# Scott Woodbury-Stewart | Founder and CEO | Scott@TargetTestPrep.com

250 REVIEWS

5-STAR RATED ONLINE GMAT QUANT SELF STUDY COURSE

NOW WITH GMAT VERBAL (BETA)

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

Current Student B
Joined: 08 Feb 2016
Posts: 67
Location: India
Concentration: Technology
Schools: AGSM '20 (A)
GMAT 1: 650 Q49 V30
GPA: 4
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

Yes, when c starts from 20, 12 numbers range from 20 to 32, inclusive

In these 12 numbers, only 3 numbers are divisible by 12. ( c=23,24 & 25)

That's how 3/12 => 1/4.
But why are we subtracting this from 1, I don't get that. DensetsuNo please explain.

Keats wrote:
DensetsuNo wrote:
$$c^{3}-c\; =\; c\left( c^{2}-1 \right)\; =\; \left( c-1 \right)c\left( c+1 \right)\; ->\; sequence\; of\; 12\; #s\; ->\; #s\; we\; have:\; 3\; ->\; 1\; -\; \frac{1}{4}\; ->\; \frac{3}{4}\; ->\; answer C$$

DensetsuNo Can you please elaborate. You say that "sequences of 12" will have 3 favorable cases? I don't know if I am interpreting it correctly. Probably with an example, I can understand better. I really want to understand this quick approach.

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

ajay2121988 wrote:
Yes, when c starts from 20, 12 numbers range from 20 to 32, inclusive

In these 12 numbers, only 3 numbers are divisible by 12. ( c=23,24 & 25)

That's how 3/12 => 1/4.
But why are we subtracting this from 1, I don't get that. DensetsuNo please explain.

Keats wrote:
DensetsuNo wrote:
$$c^{3}-c\; =\; c\left( c^{2}-1 \right)\; =\; \left( c-1 \right)c\left( c+1 \right)\; ->\; sequence\; of\; 12\; #s\; ->\; #s\; we\; have:\; 3\; ->\; 1\; -\; \frac{1}{4}\; ->\; \frac{3}{4}\; ->\; answer C$$

DensetsuNo Can you please elaborate. You say that "sequences of 12" will have 3 favorable cases? I don't know if I am interpreting it correctly. Probably with an example, I can understand better. I really want to understand this quick approach.

This is not correct. Check this link:
https://gmatclub.com/forum/if-an-intege ... l#p1746303
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 14 Mar 2017
Posts: 19
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

sameerspice wrote:
mitko20m wrote:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

the easiest way to solve this without any calculation is by as follows
1. the only way n(n+1) is divisible by 4, if and only if either n is multiple of 4, Between 1 & 100, there are 25 number that are multiple of 4, such as 4, 8, 12...100
2. or (n+1) is multiple of 4, this is equivalent to saying number which are multiple of 3, Between 1 & 100, there are 25 such as 3,7,..99,

Hence we have 25 + 25 = 50 number which can be picked between 1 and 100, that would make n(n+1) divisible by 4

probability: fav case/total occurence => 50/100 = 1/2

I think that the above statement ""r (n+1) is multiple of 4, this is equivalent to saying number which are multiple of 3"" is inaccurate because it is not necessary that such a number is a multiple of 3. For example, if n = 7, then n(n+1) = 7*8 which is divisible by 4 and 7 is not a multiple of 3.

So there are 25 numbers (n) that are multiples of 4 and therefore, there would also be 25 numbers (n) to which if 1 were added (n+1), would become multiples of 4 = 50

Therefore, answer = 50/100 = 1/2
Intern  B
Joined: 19 May 2018
Posts: 13
GPA: 3.7
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

DensetsuNo wrote:

3-sec solution:

If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

Sequence of 4 numbers
: 4
Numbers we have: 2

You can see how this method works also on harder questions letting us solve them in a few seconds.
eg. If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

Sequence of 8 numbers: 8
Numbers we have: 3

Similar questions:
http://gmatclub.com/forum/if-integer-c- ... 21561.html
http://gmatclub.com/forum/if-an-integer ... 26654.html

Why doesn't this work for the following question: https://gmatclub.com/forum/an-integer-n ... 60998.html

Or am I missing something? Btw this shortcut is beautiful just trying to better understand how to use it!
Intern  Joined: 18 Dec 2018
Posts: 45
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

n and (n + 1) are two consecutive integers.
So, one of these two integers must be odd and the other must be even.
n(n + 1) is divisible by 4 when either n or (n+ 1) is divisible by 4.
For integral values between 1 and 100(both inclusive),
n and (n + 1) each have 25 values for which it is divisible by 4. So, total no of value = 25 + 25 = 50
Probability = favorable outcome/total possible outcomes
Probability = 50/100 = 1/2.
Director  V
Joined: 06 Jan 2015
Posts: 793
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)
If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

Bunuel wrote:
mitko20m wrote:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

For n(n+1) to be a multiple of 4 either n or n+1 has to be a multiple of 4. So, n must be a multiple of 4 or 1 less than a multiple of 4:

{1, 2, 3, 4,} {5, 6, 7, 8,} {9, 10, 11, 12,} ..., {97, 98, 99, 100}.

As you can see half of the integers from 1 to 100, inclusive, guarantees divisibility by 4.

Can you please help me. How did you group as 4? Why you didn't take as a group of 3?

Suppose {1, 2, 3,}{4, 5, 6,}{ 7,8, 9}

Here if n is either 3 or 4 n(n+1) is divisible by 4
_________________
आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution

Originally posted by NandishSS on 09 Jan 2019, 06:04.
Last edited by NandishSS on 04 Feb 2019, 07:25, edited 1 time in total.
Manager  B
Joined: 12 Jan 2019
Posts: 50
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

n and (n + 1) are two consecutive integers.
So, one of these two integers must be odd and the other must be even.
n(n + 1) is divisible by 4 when either n or (n+ 1) is divisible by 4.
For integral values between 1 and 100(both inclusive),
n and (n + 1) each have 25 values for which it is divisible by 4. So, total no of value = 25 + 25 = 50
Probability = favorable outcome/total possible outcomes
Probability = 50/100 = 1/2.
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 5000
GMAT 1: 770 Q49 V46
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

Top Contributor
mitko20m wrote:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

Let's look for a pattern.

If n = 1, then (n)(n+1) = 2, which is NOT divisible by 4
If n = 2, then (n)(n+1) = 6, which is NOT divisible by 4

If n = 3, then (n)(n+1) = 12, which IS divisible by 4
If n = 4, then (n)(n+1) = 20, which IS divisible by 4

If n = 5, then (n)(n+1) = 30, which is NOT divisible by 4
If n = 6, then (n)(n+1) = 42, which is NOT divisible by 4

If n = 7, then (n)(n+1) = 56, which IS divisible by 4
If n = 8, then (n)(n+1) = 72, which IS divisible by 4

.
.
.
From the pattern, we can see that, out of every FOUR consecutive values of n, (n)(n+1) IS divisible by 4 for TWO of the values, and (n)(n+1) is NOT divisible by 4 for TWO of the values.

So, the probability is 1/2 that n(n+1) will be divisible by 4

Cheers,
Brent
_________________
If you enjoy my solutions, you'll love my GMAT prep course. Senior PS Moderator D
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 718
GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

Hi NandishSS,

We group by four for the simple reason that we are interested in the product being divisible by four. If I had changed the question to n divisible by 4 from n=1 to 100 inclusive, what would the answer be?

100/4 or 25.

Now instead of n, we have a product of n*(n+1) ... So for the product to be divisible either of them can be divisible. Also, note here that we would have to subtract the cases where both are divisible, but such a situation does not arise as they will always be co-prime a subtle but important point.

Hence 25 ways for n to be divisible and 25 ways for (n+1) to be. A total of 50 out of 100 and hence 1/2.

Hope it is clear. Let me know.

NandishSS wrote:
Bunuel wrote:
mitko20m wrote:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

For n(n+1) to be a multiple of 4 either n or n+1 has to be a multiple of 4. So, n must be a multiple of 4 or 1 less than a multiple of 4:

{1, 2, 3, 4,} {5, 6, 7, 8,} {9, 10, 11, 12,} ..., {97, 98, 99, 100}.

As you can see half of the integers from 1 to 100, inclusive, guarantees divisibility by 4.

Can you please help me. How did you group as 4? Why you didn't take as a group of 3?

Suppose {1, 2, 3,}{4, 5, 6,}{ 7,8, 9}

Here if n is either 3 or 4 n(n+1) is divisible by 4

Posted from my mobile device
_________________
Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
Director  V
Joined: 06 Jan 2015
Posts: 793
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 3?

Do you mean to if divisible by 3, then divide it by 3 i,e 100/3 to group?
_________________
आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution
Senior PS Moderator D
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 718
GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

Yes and round of correctly. So for n*(n+1) divisible by 3... It will be 33+33 or 66 out of 100.

Just check, n=2,3 n=5,6 ... So basically two out of three works.

Posted from my mobile device
_________________
Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
Senior Manager  P
Joined: 05 Feb 2018
Posts: 440
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

Very useful explanations in this post. My rewording and working through the problems:

If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

We have 100 consec ints, half are even and half are odd.
Out of the 50 even numbers, half are divisible by 4 (can check quickly by writing out: {1,2,3,4} {2,3,4,5} {3,4,5,6} {4,5,6,7} or use the formula
Last multiple of 4 (100) - First multiple of 4 (4) / 4 + 1 = 24 + 1 = 25.
BUT, this is the pattern for n alone. If we have n(n+1) that means either n or n+1 can be one out of 4 numbers divisible by 4. So 25*2 --> 50/100 = 1/2.

In other words, we have a repeating pattern of 4 numbers where 1n out of 4n is divisible by 4. {1,2,3,4} choose 1, 2, 3 or 4.
This pattern is repeated 25 times out of 100 from 1-100 {1,2,3,4} ... {97,98,99,100} when we have 1 choice.
So we have 25/100 = 1/4 choices for just n.

For n(n+1) we have 2 choices from {1,2,3,4} choose 1,2 or 2,3 or 3,4, 4,5
The product of 2 out of those 4 is divisible by 8. That means 50 out of 100 possible numbers or 1/2.

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

Given: 96 consec ints from 1-96
1) Out of every 8 consecutive ints, for just n have 1/8 that is acceptable.
2) But, we have 3 choices to make a number divisible by 8, n(n+1)(n+2)

Unlike the previous example, there isn't symmetry since there's 2 YES cases : Where n(n+1)(n+2) is E#O#E#, and when n(n+1)(n+2) is O#E#O# AND n+1 = a multiple of 8. The single NO case is when O#E#O# and the E# does not have enough factors of 2.

From #1-8 we can pick and see which 3 numbers have enough factors of 2 to make the product divisible by 8.
n,n+1,n+2
{1,2,3} no
{2,3,4} yes
{3,4,5} no
{4,5,6} yes
{5,6,7} no
{6,7,8} yes
{7,8,9}yes
{8,9,10} yes

It is symmetrical (Case 1 E#O#E#) until we get to {7,8,9} which is our yes Case 2, where even though n = O#, we have enough factors of 2 in the E# n+1.

So every 8 numbers, we have 5/8 choices where it is divisible by 8.

If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?
Given: 99-20+1 = 80 consec ints from 20 to 99.
1) Out of every 12 ints, 1 is divisible by 12
2) We have (n-1)n(n+2) so it is slightly different from the problem above.

Question: do the 3 numbers have 2 factors of 2? (3 consec ints are always divisible by 3, but may not be divisible by 4)
We have 2 Yes cases (all E#O#E# are at least 2 factors of 2, and O#E#O# where n has at least 2 factors of 2). The No case is when O#E#O# and n doesn't have enough factors of 2.

Sequence of 12 numbers where n = 20 through 31
n-1, n, n+1
{19,20,21} yes, case 2
{20,21,22} yes, case 1
{21,22,23} no
{22,23,24} yes, case 1
{23,24,25} yes, case 2
{24,25,26} yes, case 1
{25,26,27} no
{26,27,28} yes, case 1
{27,28,29} yes, case 2
{28,29,30} yes, case 1
{29,30,31} no
{30,31,32} yes, case 1

So 9/12 or cases are YES, 3/12 are NO.

VeritasKarishma I hope that is all correct, my brain is fried from typing all that out.
I don't understand the logic behind how 1 - number of choices/sequence gives us the right answer all 3 times here. I can understand finding the probability from 1-Not A = A but how does Not A relate to number of choices/sequence? Can you give an example of when this shortcut doesn't work?
BSchool Moderator S
Joined: 29 Apr 2019
Posts: 96
Location: India
Re: If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

### Show Tags

Hi,

I'd really appreciate if someone can solve this in a more intuitive way:

P [n(n+1) div 4] = {number of multiples of 3 between 1 and 100 + number of multiples of 4 - common multiples between 3 and 4} / 100

Is this the correct way?

I'm getting 24 + 33 - 8. where am i going wrong? Re: If an integer n is to be selected at random from 1 to 100, inclusive,   [#permalink] 30 Aug 2019, 05:01

Go to page   Previous    1   2   [ 38 posts ]

# If an integer n is to be selected at random from 1 to 100, inclusive,  