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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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New post 10 Oct 2016, 04:50
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

Answer 1-3/8= 5/8.


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?

Please help me understand. Many thanks in advance.


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]

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New post 10 Oct 2016, 05:11
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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New post 10 Oct 2016, 08:10
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

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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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New post 10 Oct 2016, 23:12
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
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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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New post 11 Oct 2016, 16:02
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

Answer: C
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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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New post 15 Feb 2017, 02:45
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.

Thank you in advance.
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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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New post 15 Feb 2017, 04:00
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.

Thank you in advance.



This is not correct. Check this link:
https://gmatclub.com/forum/if-an-intege ... l#p1746303
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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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New post 09 Apr 2018, 23:54
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
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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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New post 24 Aug 2018, 23:05
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

Answer 1-3/8= 5/8.

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!
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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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New post 18 Dec 2018, 22:09
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.
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If an integer n is to be selected at random from 1 to 100, inclusive,  [#permalink]

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New post Updated on: 04 Feb 2019, 08:25
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.

Answer: C.



HI Bunuel, VeritasKarishma, Gladiator59, generis

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
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Originally posted by NandishSS on 09 Jan 2019, 07:04.
Last edited by NandishSS on 04 Feb 2019, 08:25, edited 1 time in total.
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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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New post 12 Jan 2019, 22:18
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.
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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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New post 28 Jan 2019, 07:08
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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

Answer: C

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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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New post 04 Feb 2019, 10:59
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.

Answer: C.



HI Bunuel, VeritasKarishma, Gladiator59, generis

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


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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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New post 04 Feb 2019, 20:14
HI Gladiator59,

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?
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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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New post 04 Feb 2019, 22:45
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.

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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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New post 11 Feb 2019, 11:32
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?
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