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

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Veritas Prep GMAT Instructor
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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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10 Oct 2016, 04:50
1
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]

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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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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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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

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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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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.

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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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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.

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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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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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

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!
Re: If an integer n is to be selected at random from 1 to 100, inclusive, &nbs [#permalink] 24 Aug 2018, 23:05

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