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If an integer n is to be chosen at random from the integers

This topic has 6 expert replies and 1 member reply

If an integer n is to be chosen at random from the integers

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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

OAD

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Quote:
If an integer n to be chosen randomly between 1 and 96 inclusive, what is the probability that n(n+1)(n+2) is divisible by 8 ?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

OA: D
First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D

Cheers,
Brent

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GMATsid2016 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

OAD
Well, we know the denominator here is 96, so we just have to figure out how many ways we can multiply three consecutive numbers to get a multiple of 8. Put another way, any multiple of 8, or 2^3, must contain three 2's. One way this can happen is if the middle number is odd, because every odd number must be sandwiched between a multiple of 2 and a multiple of 4. If n+1 is 3, for example, you'd have 2*3*4 --> multiple of 8. If n+1 is 5, you'd have 4*5*6 --> multiple of 8. Between 1 and 96, we've got 48 odds.

The other way we can get a multiple of 8 is if n + 1 is itself a multiple of 8. 96/8 = 12 multiples of 8 between 1 and 96.

So we've got 48 odds + 12 multiples of 8 for a total of 60 desired outcomes.

60/96 = 5/8. Answer is D

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Hi Sid,

Brent's already listed out the pattern in the sequence, so I won't rehash any of that here. Instead, I'll focus on WHY that pattern exists.

For a number to be evenly divisible by 8, it has to include at least three 2's when you prime factor it.

For example,
8 is divisible by 8 because 8 = (2)(2)(2).....it has three 2s "in it"
48 is divisible by 8 because 48 = (3)(2)(2)(2)(2).....it has three 2s "in it" (and some other numbers too).

20 is NOT divisibly by 8 because 20 = (2)(2)(5)....it only has two 2s.

In this question, when you take the product of 3 CONSECUTIVE POSITIVE INTEGERS, you will either have....

(Even)(Odd)(Even)

or

(Odd)(Even)(Odd)

In the first option, you'll ALWAYS have three 2s. In the second option, you'll only have three 2s if the even term is a multiple of 8 (Brent's list proves both points). So for every 8 consecutive sets of possibilities, 4 of 4 from the first option and 1 of 4 from the second option will give us multiples of 8. That's 5/8 in total.

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Another approach:

If n is even, then (n + 2) is also even. If we multiply two consecutive even integers together, the result is always divisible by 8: 2*4, 4*6, 6*8, 8*10, etc.

So if n is even, we're set. That gets us to AT LEAST 1/2. Eliminate A, B, and C.

Now we want to see what happens if n is odd. Try a few:

n = 1, nope
n = 3, nope
n = 5, nope
n = 7, yup

So it looks like this will work about 1/4 of the time. n is odd 1/2 the time, and 1/4 of the odds seem to work, so this is another (1/2) * (1/4), or 1/8.

Even cases + Odd cases = 1/2 + 1/8 = 5/8, so we're set.

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GMATsid2016 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

OAD
We are given that an integer n is to be chosen at random from the integers 1 to 96 inclusive, and we need to determine the probability that n(n + 1)(n + 2) will be divisible by 8.

We should recall that when a number is divisible by 8, it is divisible by 2^3, i.e., three factors of 2. We should also recognize that n(n + 1)(n + 2) is the product of three consecutive integers.

Case 1: n is even

Any time that n is even, n + 2 will also be even. Moreover, either n or n + 2 will be divisible by 4, and thus n(n + 1)(n + 2) will contain three factors of 2 and will be divisible by 8.

Since there are 96 integers between 1 and 96, inclusive, and half of those integers are even, there are 48 even integers (i.e., 2, 4, 6, …, 96) from 1 to 96 inclusive. Thus, when n is even, there are 48 instances in which n(n + 1)(n + 2) will be divisible by 8.

Case 2: n is odd

If n is odd, then n(n + 1)(n + 2) still can be divisible by 8, but only if the factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 96, inclusive.

Number of multiples of 8 = (96 - 8)/8 + 1 = 88/8 + 1 = 12. Thus, when n is odd, there are 12 instances in which n(n + 1)(n + 2) will be divisible by 8.

In total, there are 48 + 12 = 60 outcomes in which n(n + 1)(n + 2) will be divisible by 8.

Thus, the probability that n(n + 1)(n + 2) is divisible by 8 is: 60/96 = 10/16 = 5/8.

Answer:D

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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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GMATsid2016 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
$$?\,\,\, = \,\,\,{{{?_{\,{\rm{temp}}}}\,\,\,\,\left( {{\rm{favorable}}\,\,{\rm{remainders}}} \right)} \over {8\,\,\,\,\left( {{\rm{equiprobable}}\,\,{\rm{remainders}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{division}}\,\,{\rm{by}}\,\,{\rm{8}}} \right)}}$$
$$\left. \matrix{
n\,\, \in \,\,\left\{ {8M,8M + 2,8M + 4,8M + 6} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,n\;\;{\rm{even}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {n + 1} \right)\left( {n + 2} \right)} \over 8} = {\mathop{\rm int}} \,\,\,\, \hfill \cr
n\,\, \in \,\,\left\{ {8M + 1,8M + 3,8M + 5} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {n + 1} \right)\left( {n + 2} \right)} \over 8} \ne {\mathop{\rm int}} \hfill \cr
n\,\, \in \,\,\left\{ {8M + 7} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {8M + 8} \right)\left( {n + 2} \right)} \over 8} = {\mathop{\rm int}} \hfill \cr} \right\}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{?_{\,{\rm{temp}}}} = 5$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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