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

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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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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
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because n(n+1) is always an even product of even*odd or odd*even factors,
there is a probability of 1 that that it will be divisible by 2,
and, thus, a probability of 1/2 that it will be divisible by 4
1*1/2=1/2
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If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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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,} ..., {87, 98, 99, 100}.

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

Thanks for your great solution.

Can you elaborate probability when n must be a multiple of 4?
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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AbdurRakib 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,} ..., {87, 98, 99, 100}.

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

Thanks for your great solution.

Can you elaborate probability when n must be a multiple of 4?

As from the highlighted list, every third and fourth digit is a multiple of 4, total no. of multiple till 100 would be 50. i.e 25 for n + 25 for (n+1)
Probability = Number of desired or successful outcomes/ Total number of possible outcomes
Probability = 50/ 100 = 1/2
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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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 divisible by 4, either n or n+1 must be divisible by 4

In 100 integers, number of integers divided by 4 are 100/4= 25.

Probability of selecting these 25 numbers= 25/100= 1/4----------------------------------- 1

for n+1 to be divisible by 4, n can be 3, 7, 11.... Basically in the form of 3+4x. The last integer in the form of 3+4x is 3+ 4*24= 99

We see that there are 24 numbers that can be written in the form of 3+4x (where x is positive integer) and 3 that will give n+1 divisible by 4. Total of 25 numbers that can be written in teh form of n+1, which will be divisible by 4

Probability of selecting these numbers= 25/100= 1/4 ------------------------------------------ 2

Total probability will be= 1/4 + 1/4= 1/2

C is the answer
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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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

I am not sure what it is that you are looking for. Bunuel has already provided the solution. I can give it in my own words, if that helps.

One number out of n and (n+1) will be odd and the other will be even. So for n(n+1) to be divisible by 4, the even number must be a multiple of 4. The other number will be odd only so it will have no multiple of 2 or 4 etc.

So, n(n+1) will be divisible by 4 if either n or (n+1) is divisible by 4.
Consider the given range:

n*(n+1) = 1*2 (n = 4a + 1)
n*(n+1) = 2*3 (n = 4a + 2)
n*(n+1) = 3*4 (n = 4a + 3, n+1 = 4b)
n*(n+1) = 4*5 (n = 4b)
n*(n+1) = 5*6 (n = 4b + 1)
n*(n+1) = 6*7 (n = 4b + 2)
n*(n+1) = 7*8 (n = 4b + 3, n+1 = 4c)
n*(n+1) = 8*9 (n = 4c)
and so on...

So of every 4 values of n, 2 will have n*(n+1) a multiple of 4 and 2 will not.
The total possible values of n are 100, a multiple of 4.
So half the values will have n(n+1) a multiple of 4, and half will not.

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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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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.
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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!
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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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
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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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

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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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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
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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.

Hi,
Does the denominator for the sets (here- four numbers- 1,2,3,4) always be equal to the number whose multiple we are checking? (here 4)
Here the last number is divisible by 4, thus sets of 4 can be made. What if n had to be selected from a the set 3 to 97 both inclusive?
Does the GMAT ask this type of question where the range cannot be divided into sets with number of elemnts equal to the number for which we have to check the divisibility?

Looking forward to hearing from you and CrackverbalGMAT KarishmaB JeffTargetTestPrep and others
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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RenB 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.

Hi,
Does the denominator for the sets (here- four numbers- 1,2,3,4) always be equal to the number whose multiple we are checking? (here 4)
Here the last number is divisible by 4, thus sets of 4 can be made. What if n had to be selected from a the set 3 to 97 both inclusive?
Does the GMAT ask this type of question where the range cannot be divided into sets with number of elemnts equal to the number for which we have to check the divisibility?

Looking forward to hearing from you and CrackverbalGMAT KarishmaB JeffTargetTestPrep and others

If the numbers were in the range 3 to 97,
the first multiple of 4 would be 4 = 1 * 4 and last would be 96 = 4 * 24 so you would get 24 such numbers. n could any of these 24 values.
Also, n could take values 1 less than these multiples i.e. 3 or 7 or ... 95 i.e. another 24 values.

So n could take 48 values such that n(n+1) is a multiple of 4.
Total numbers are 97 - 3 + 1 = 95

Probability = 48/95
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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

But there are 33 multiples of 3 between 1 and 100 inclusive... (99-3)/3 +1 = 33.
I thought that this approach worked but now I'm confused because 33+25 does not equal 50
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
josephinedemetz wrote:
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

But there are 33 multiples of 3 between 1 and 100 inclusive... (99-3)/3 +1 = 33.
I thought that this approach worked but now I'm confused because 33+25 does not equal 50

The reasoning in that solution isn't accurate. The assertion that if n + 1 is a multiple of 4, then n has to be a multiple of 3 is not necessarily correct. For instance, when considering n + 1 = 8, which is indeed a multiple of 4, n = 7 isn't a multiple of 3.

However, if n + 1 is a multiple of 4, then n is 1 less than a multiple of 4. From 1 to 100, inclusive, there are 25 numbers that are one less than a multiple of 4, and an equal number, 25, that are multiples of 4. Hence, the probability is (25 + 25)/100, which simplifies to 1/2.
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
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

Answer: 1-2/4 = 1/2

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:
https://gmatclub.com/forum/if-integer-c ... 21561.html
https://gmatclub.com/forum/if-an-intege ... 26654.html

Hey! How would you go about the second questions that you posted under similar question? Also what if they are exclusive?
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Re: If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]
We need the multiples of 4 and (4+1). The pattern of multiples of 4 are:
From 1-10 = 2 multiples
From 11-2+ 3 multiples and this contunies 2+3+2+3+2+ 2+3+2+3+2+ . Therefore, total multiples of 4 in 1-100= 25.

n(n+1) = 25+ 25 =50
Total number= 100

50/100= 1/2.