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

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.
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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:
if-integer-c-is-randomly-selected-from-20-to-99-inclusive-121561.html
if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html

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

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

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

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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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.

Answer (C)
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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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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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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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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

Cheers,
Brent
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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?

Did not understand how we got the 'numbers we have' section

This doesnt work though if we have, for example, 4 numbers and want to divise by 12.

1-4/12 = 8/12

But in fact every string of 4 numbers will be divisible by 12, because at least one number will be divisible by 4 and at least one divisible by 3.

Hi Brent,
I am just wondering for a fail-safe mechanism in case things go wrong with other approaches to these types of problems. Does a pattern approach always work irrespective of the denominator? e.g. 3,4,5,8,12, etc.? And will the numerator ever be something beyond consecutive integers: e.g. n x (n+1) x (n + 2) or (n-1) x n x (n + 1)?
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I think it is not accurate to say that we need to find "numbers which are multiple of 3". If we look for numbers that are multiple of 3, then the value 7 won't be in that list and total q-ty of numbers from 1 to 100 that are multiple of 3 will be 33.

Hi DensetsuNo thanks for the fast solution, but I got a bit unclear about something as below. Will appreciate if you would help explain.

in a sequence of 4 numbers we have 1,2,3,4; in this sequence n could be 3 or 4 based on the divisibility criteria. Hence probability of getting picked is 2/4, which is 1/2. When we are subtracting 1/2 from 1, I am assuming here that in any # of 4 digit sequence (in our case there are 25 sets making 100 total), the probability of picking multiples of 4 = probability of picking any numbers - probability of picking non multiples of 4 ===> 1 - 1/2 = 1/2

In the second example you have up there, I am finding it difficult to understand the 1-3/8 subtraction. Would you please help elaborate a bit on this?