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

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

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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.

Answer: C.

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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New post 16 Jul 2016, 23:27
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.

Answer: C.

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

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Dear Bunuel ,I meant probability when n=4,8,or 12 etc,is that possible?

Wording of the question is not clear to me
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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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If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]

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Bunuel wrote:
AbdurRakib wrote:
Dear Bunuel ,I meant probability when n=4,8,or 12 etc,is that possible?

Wording of the question is not clear to me


The question asks about the probability that the product of n and (n+1) is divisible by 4.


Sorry to ask

Is that possible the pair n(n+1) could be (4,5),(8,9) or (12,13) etc.?
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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 17 Jul 2016, 01:44
AbdurRakib wrote:
Bunuel wrote:
AbdurRakib wrote:
Dear Bunuel ,I meant probability when n=4,8,or 12 etc,is that possible?

Wording of the question is not clear to me


The question asks about the probability that the product of n and (n+1) is divisible by 4.


Sorry to ask

Is that possible the pair n(n+1) could be (4,5),(8,9) or (12,13) etc.?



Yes, the answer could be calculated in this way but it is time consuming. The best way to solve this question is to find out the number of multiples of 4 in 1-100 and then multiple the result by 2 as each multiple will make two pairs of n(n+1) such that it is divisible by 4.

100= 4+(n-1)*4

=> n=25.

Therefore number of such pairs would be 25*2=50.

For example, we could have a pair of (3,4) and (4,5); (7,8) and (8,9), etc.

Hence probability would be 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 17 Jul 2016, 02:43
AbdurRakib wrote:
Bunuel wrote:
AbdurRakib wrote:
Dear Bunuel ,I meant probability when n=4,8,or 12 etc,is that possible?

Wording of the question is not clear to me


The question asks about the probability that the product of n and (n+1) is divisible by 4.


Sorry to ask

Is that possible the pair n(n+1) could be (4,5),(8,9) or (12,13) etc.?


Yes it is possible as the pairs (4,5),(8,9) or (12,13) etc. are subset of the 50 possibilities we found.
25 in which the first digit n is divisible by 4 - (4,5),(8,9) or (12,13)
and another 25 in which the second digit (n+1) is divisible by 4 (3,4), (7,8), (11,12)

Hope i understood the question properly and this clears the query.
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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 17 Jul 2016, 06:40
Thanks Bunuel! I guess I was confused by the n(n+1). I thought the product n(n+1) should be divisible by 4, and I couldn't quickly think of when
that would be the case. It didn't occur to me that we should think of it as "either "n" or "(n+1)". It would have been clearer if the question explicitly stated "n" or n+1..
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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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If an integer n is to be selected at random from 1 to 100, inclusive, [#permalink]

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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
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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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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:
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


Is there a kind of rule or theory?
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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 20 Sep 2016, 08:15
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.

Answer (C)
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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 09 Oct 2016, 00:27
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:
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


Why do you subtract with 1 here? In both the cases? Also, can you apply your method to below question -

If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?

A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3
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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 Oct 2016, 00:52
Keats wrote:
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:
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


Why do you subtract with 1 here? In both the cases? Also, can you apply your method to below question -

If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?

A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3


\(c^{3}-c\; =\; c\left( c^{2}-1 \right)\; =\; \left( c-1 \right)c\left( c+1 \right)\; ->\; sequence\; of\; 12\; #s\; ->\; #s\; we\; have:\; 3\; ->\; 1\; -\; \frac{1}{4}\; ->\; \frac{3}{4}\; ->\; answer C\)
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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 09 Oct 2016, 01:45
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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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, 01:15
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


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.
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