integers and prob : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 12:32

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# integers and prob

Author Message
TAGS:

### Hide Tags

VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 8

Kudos [?]: 484 [1] , given: 0

### Show Tags

19 Jan 2008, 05:19
1
KUDOS
00:00

Difficulty:

45% (medium)

Question Stats:

60% (01:41) correct 40% (02:33) wrong based on 10 sessions

### HideShow timer Statistics

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?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
[Reveal] Spoiler: OA
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 545

Kudos [?]: 3551 [0], given: 360

### Show Tags

19 Jan 2008, 12:18
B

take a look at integer sequence:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19

1,2,3,4,5,[6,7,8],9,10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,[7,8,9],10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,7,[8,9,10],11,12,13,14,15,16,17,18,19

Therefore, we have 3 combinations for each 8 consecutive integers.
$$p=\frac38$$
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 19

Kudos [?]: 420 [0], given: 0

### Show Tags

19 Jan 2008, 12:33
walker wrote:
B

take a look at integer sequence:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19

1,2,3,4,5,[6,7,8],9,10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,[7,8,9],10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,7,[8,9,10],11,12,13,14,15,16,17,18,19

Therefore, we have 3 combinations for each 8 consecutive integers.
$$p=\frac38$$

Why do we have to start at 8?
if n=2 then
2(3)(4) is divisible by 8?

I get every even number from 1-96 when n(n+1)(n+2) is divisible by 8.
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 545

Kudos [?]: 3551 [0], given: 360

### Show Tags

19 Jan 2008, 12:41
ah... you are right

D

take a look at integer sequence:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19

1,2,3,4,5,[6,7,8],9,10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,[7,8,9],10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,7,[8,9,10],11,12,13,14,15,16,17,18,19

Therefore, we have 3 combinations for each 8 consecutive integers.

to catch 8*n we also have to start with other even integers 8n-6, 8n-4

1,[2,3,4],5,6,7,8,9,10,11,12,13,14,15,16,17,18,19
1,2,3,[4,5,6],7,8,9,10,11,12,13,14,15,16,17,18,19

$$p=\frac38+\frac28=\frac58$$
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 545

Kudos [?]: 3551 [0], given: 360

### Show Tags

19 Jan 2008, 12:43
GMATBLACKBELT wrote:
I get every even number from 1-96 when n(n+1)(n+2) is divisible by 8.

plus integers like (7,8,9) or (15,16,17)....so 4/8+1/8=5/8
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 03 Sep 2006
Posts: 879
Followers: 6

Kudos [?]: 768 [0], given: 33

### Show Tags

19 Jan 2008, 22:01
marcodonzelli wrote:
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?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

1,2,3,...96.

n can be chosen in 96 ways. ( This is total number of samples )

$$n*(n+1)*(n+2)/8$$ implies we should find number of favorable samples in which we have

$$n*(n+1)*(n+2) = 8(min), 16, 32, ...96(max)$$

$$8 = 2^3$$ ( Thus total number of divisors of 8 will be $$(3+1)=4$$ ) which are

1, 2, 4, 8. From this we get 3 different possibilities;

$$2*(2+1)*(2+2)$$
$$4*(4+1)*(4+2)$$
$$8*(8+1)*(8+2)$$

But with 8 we can also have;

$$6*(6+1)*(6+2)$$
$$7*(7+1)*(7+2)$$

Similarly for $$16 = 2^4$$ ( Thus total number of divisors of 8 will be $$(4+1)=5$$ ) which are

1, 2, 4, 8, 16. From this also we get 4 different possibilities;

$$2*(2+1)*(2+2)$$
$$4*(4+1)*(4+2)$$
$$8*(8+1)*(8+2)$$ Two more associated with it as shown above.
$$16(16+1)*(16+2)$$ Two more associated with it depending on the position of 16.

Two of them are already in the sequence, so 3 different.

and so on... with each multiple of 8, we will get 3 different possibilities.

Now, how many are the total number of number which are multiple of 8 from 8...96.
$$(96-8/8)+1=12$$

Each of these 12 different multiples have 3 different possibilities in which 8 can be the divisor.

Thus total number of favorable outcomes = $$12*3 + 2*(2+1)*(2+2) + 4*(4+1)*(4+2) = 36+1+1$$

$$P = 38/96$$

Did I miss something????
Manager
Joined: 04 Jan 2008
Posts: 84
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

24 Jan 2008, 10:43
A no can be divided by eight if we can divide that no by both 2 and 4.
n(n+1)(n+2) , three consecutive nos can always be divided by 2,
and if n is even, we can divide by 4 also
Therefore for all even n we can divide n(n+1)(n+2) by 8
we have 48 even and 96 total nos
so, prob = 48/96= 1/2
whats the OA pls?????
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 545

Kudos [?]: 3551 [0], given: 360

### Show Tags

24 Jan 2008, 10:49
Dellin wrote:
A no can be divided by eight if we can divide that no by both 2 and 4.
n(n+1)(n+2) , three consecutive nos can always be divided by 2,
and if n is even, we can divide by 4 also
Therefore for all even n we can divide n(n+1)(n+2) by 8
we have 48 even and 96 total nos
so, prob = 48/96= 1/2

plus integers like (7,8,9) or (15,16,17)....so 1/2+1/8=5/8
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 8

Kudos [?]: 484 [0], given: 0

### Show Tags

20 Feb 2008, 00:46
walker wrote:
Dellin wrote:
A no can be divided by eight if we can divide that no by both 2 and 4.
n(n+1)(n+2) , three consecutive nos can always be divided by 2,
and if n is even, we can divide by 4 also
Therefore for all even n we can divide n(n+1)(n+2) by 8
we have 48 even and 96 total nos
so, prob = 48/96= 1/2

plus integers like (7,8,9) or (15,16,17)....so 1/2+1/8=5/8

walker can you help me with this approach?

96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss?
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 545

Kudos [?]: 3551 [0], given: 360

### Show Tags

20 Feb 2008, 01:35
marcodonzelli wrote:
walker can you help me with this approach?

96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss?

I think the approach is not correct. You missed the fact that we have 3 consecutive integers. For example, 4,5,6 are not multiplies of 8 but 4*5*6 is divisible by 8.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 01 Jan 2008
Posts: 629
Followers: 4

Kudos [?]: 175 [0], given: 1

### Show Tags

20 Feb 2008, 09:20
marcodonzelli wrote:
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?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

Total numbers 8*12
There are 12 numbers divisible by 8 -> 3*12 (if 8 is an example - (6,7,8), (7,8,9), (8,9,10))
and 12 numbers divisible by 4 but not divisible by 8 -> 2*12 (if 4 is an example (2,3,4) and (4,5,6))

Director
Joined: 31 Mar 2007
Posts: 585
Followers: 9

Kudos [?]: 64 [1] , given: 0

### Show Tags

20 Feb 2008, 11:24
1
KUDOS

if n is even, then n*n+2 is divisible by 8. so we have boom 50% so far.

then, we consider if n+1 is divislbe by 8, which is another 12 times

so 48 + 12 / 96 = 60/96 = 30/48 = 5/8

QED
VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 8

Kudos [?]: 484 [0], given: 0

### Show Tags

20 Feb 2008, 22:56

if n is even, then n*n+2 is divisible by 8. so we have boom 50% so far.

then, we consider if n+1 is divislbe by 8, which is another 12 times

so 48 + 12 / 96 = 60/96 = 30/48 = 5/8

QED

that's a great job really. quick and easy. kudo for you
VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 8

Kudos [?]: 484 [1] , given: 0

### Show Tags

17 Mar 2008, 02:32
1
KUDOS
marcodonzelli wrote:

if n is even, then n*n+2 is divisible by 8. so we have boom 50% so far.

then, we consider if n+1 is divislbe by 8, which is another 12 times

so 48 + 12 / 96 = 60/96 = 30/48 = 5/8

QED

that's a great job really. quick and easy. kudo for you

I didn't imagine it could be so easy, without any use of calculus and without trying with numbers, somewhat very difficult to do in 2 mins during the exam:

96/8=12 numbers divisible by 8

(12 + 12 + 12)/96=3/8. it took me 1 second
Intern
Joined: 20 Apr 2008
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

28 Apr 2008, 00:09
OK, here is a long explanation for a problem that took me less than a minute.

First and most important, we know that for n(n+1)(n+2) to be divisible by 8, either n or n+1 or n+2 must be divisible by 8

Second we have to realize that we are dealing with consecutive integers.

We also realize that as we have consecutive integers and multiplication, every even integer n from 2 to 96 satisfy n(n+1)(n+2) to be divisible by 8, allowing us to quickly count all even integers from 2 to 96 to be 47, when we look when n or n+1 or n+2 is divisible by 8.

Realizing the above three in a snap, we simply check when n or n+1 or n+2 is divisible by 8:

1. n is divisible by 8 when n is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8.

2. n+1 is divisible by 8, when n+1 is 8,16,32,36,40,48, 56,64,72,80,88,96, but as we deal with consecutive integers, this means that in such case, n will be odd integer 7,31,35,39,47,55,63,71,79,87,95, therefore we can pick another 13 integers that make n(n+1)(n+2) divisible by eight. We have to add those 13 odds to our 47 evens.

3. n+2 is divisible by 8, when n+2 is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8.

we have 47+13 integers satisfying n(n+1)(n+2) is divisible by 8, thus the probability of picking one of these integers is simply P(E) = n of event occuring/total number, that is 60/96, which is equal to 5/8.
SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 34

Kudos [?]: 861 [0], given: 5

### Show Tags

24 Aug 2008, 13:15
marcodonzelli wrote:
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?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

good question.

when N is even.. n(n+2) is divisable by 8
when n is odd.. n+1 must be divisible by 8 to make n(n + 1)(n + 2)

probability = (96/2 + 96/8 )/96 = 60/96= 5/8
_________________

Smiling wins more friends than frowning

SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 514 [0], given: 36

### Show Tags

31 Mar 2011, 00:25
The OA is wrong here, it should be D.
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 161

Kudos [?]: 1702 [0], given: 376

### Show Tags

31 Mar 2011, 00:31
subhashghosh wrote:
The OA is wrong here, it should be D.

Thanks. Corrected.
_________________
Re: Divisibility by 8   [#permalink] 31 Mar 2011, 00:31
Similar topics Replies Last post
Similar
Topics:
1 Of a set of 25 consecutive integers beginning with 4, what is the prob 6 13 Dec 2016, 06:06
Prob solving 2 08 Jun 2011, 08:32
2 Polygon prob 5 31 Jan 2011, 13:57
Plane prob 3 18 Oct 2010, 19:59
1 prob. wordings 3 08 Jul 2010, 03:55
Display posts from previous: Sort by