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Tough Probability Q

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by GMATGuruNY » Tue Jun 16, 2015 11:22 am
prachi18oct wrote:If 1 ≤ x ≤ 100, what is the probability that x(x + 1) is a multiple of 12 and 9?

A)1/25
B)1/20
C)2/25
D)11/100
E)1
P = (good options)/(all possible options).

Since 1 ≤ x ≤ 100, there are 100 possible options for x(x+1).

For x(x+1) to be a multiple of 9, either x or x+1 must be a multiple of 9.
For x(x+1) to be a multiple of 12, either x or x+1 must be a multiple of 4.

Make a list of options for x(x+1) in which either x or x+1 is a multiple of 9 and either x or x+1 is a multiple of 4:
8*9
27*28
35*36
36*37
44*45
63*64
71*72
72*73
80*81
99*100
Total good options = 10.

Thus:
P = (good options)/(all possible options) = 10/100 = 1/10.

The correct answer does not seem to be among the answer choices.
What is the source?
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by prachi18oct » Tue Jun 16, 2015 11:36 am
GMATGuruNY wrote:
prachi18oct wrote:If 1 ≤ x ≤ 100, what is the probability that x(x + 1) is a multiple of 12 and 9?

A)1/25
B)1/20
C)2/25
D)11/100
E)1
P = (good options)/(all possible options).

Since 1 ≤ x ≤ 100, there are 100 possible options for x(x+1).

For x(x+1) to be a multiple of 9, either x or x+1 must be a multiple of 9.
For x(x+1) to be a multiple of 12, either x or x+1 must be a multiple of 4.

Make a list of options for x(x+1) in which either x or x+1 is a multiple of 9 and either x or x+1 is a multiple of 4:
8*9
27*28
35*36
36*37
44*45
52*54
63*64
71*72
72*73
80*81
99*100
Total good options = 11.

Thus:
P = (good options)/(all possible options) = 11/100.

The correct answer is D.
Apparently, the OA is ambigious.
Here is the official explanation.

Any number that is a multiple of 12 and 9 is a multiple of the least common multiple of 12 and 9. The least common multiple is 36. Thus the question is what is the probability that x(x + 1) is a multiple of 36.

In order for x(x + 1) to be a multiple 36, x must be a multiple of 36, x + 1 must be a multiple of 36 (which means that x would be one less than a multiple of 36), OR x multiplied x + 1 must be a multiple of 36.

Thus the question is what is the probability that x is either a multiple of 36 or 1 less than a multiple of 36 or that both x and x + 1 have 36 as their least common multiple.

Since there are 2 multiples of 36 from 1 - 100, inclusive, 36 and 72, there will also be 2 numbers which are one less than a multiple of 36, 35 and 71. The are no two consecutive integers that can multiply together to yield 36. But we find that 8 and 9 have 72 as a product.

The total number of integers that will make x(x + 1) divisible by 36 from 1 - 100 is 5.

Since the total number of integers in question is 100, put the number that meet the requirement over the total number.
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by Ian Stewart » Tue Jun 16, 2015 12:02 pm
This is a really tedious question - we want values of x that make x(x+1) divisible by 2^2 * 3^2. Since we're multiplying consecutive integers, we're multiplying two numbers with a GCD of 1, so only one of them can possibly be divisible by 3, and we know then that one of x or x+1 is a multiple of 9. By listing multiples of 9, you can count how many work. I get the following list:

8*9
27*28
35*36
36*37
44*45
63*64
71*72
72*73
80*81
99*100

of ten possible values of x, which makes the answer 10/100 = 1/10, which isn't among the answer choices. The "official" explanation is simply wrong, so it's not surprising that the question doesn't have the right answer as one of the choices.
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by MartyMurray » Tue Jun 16, 2015 12:30 pm
prachi18oct wrote:
Apparently, the OA is ambigious.
Here is the official explanation.

Any number that is a multiple of 12 and 9 is a multiple of the least common multiple of 12 and 9. The least common multiple is 36. Thus the question is what is the probability that x(x + 1) is a multiple of 36.

In order for x(x + 1) to be a multiple 36, x must be a multiple of 36, x + 1 must be a multiple of 36 (which means that x would be one less than a multiple of 36), OR x multiplied x + 1 must be a multiple of 36.

Thus the question is what is the probability that x is either a multiple of 36 or 1 less than a multiple of 36 or that both x and x + 1 have 36 as their least common multiple.

Since there are 2 multiples of 36 from 1 - 100, inclusive, 36 and 72, there will also be 2 numbers which are one less than a multiple of 36, 35 and 71. The are no two consecutive integers that can multiply together to yield 36. But we find that 8 and 9 have 72 as a product.

The total number of integers that will make x(x + 1) divisible by 36 from 1 - 100 is 5.

Since the total number of integers in question is 100, put the number that meet the requirement over the total number.
I took the question to mean that x(x + 1) is a multiple of both 12 and 9.

The first step that jumps out at me is breaking down 12 and 9 into their prime factors.

9 = 3 x 3

12 = 3 x 2 x 2.

So anything that is a multiple of both 12 and 9 will have as factors at least 2, 2, 3, and 3.

The smallest such number is 2 x 2 x 3 X 3 = 36. Any multiple of 36 will also work.

So the question becomes out of the 100 integers from 1 to 100 for how many integers, x, does x(x + 1) multiply to 36 or a multiple of 36.

There is no way to get the prime factors of 36, 2, 2, 3, and 3, to become x(x + 1).

However, 35 x 36 and 36 x 37 are multiples of 36. So 35 and 36 are two possible values of x.

72 is also a multiple of 36. So 71 and 72 also work as x such that x(x + 1) is a multiple of 36.

So we have the four easy ones. The next multiple of 36 is 108, and both 107 and 108 are out of the range of 1 to 100.

Now that we got the easy ones, we have to see how many ways the prime factors of 36 can be used another way to get to an integer x such that x(x + 1) is a multiple of 36.

We need to get two 3's and two 2's somehow out of x(x + 1). Even numbers alternate with odd numbers. So basically we need multiples of 4 followed by multiples of 9 or multiples of 9 followed by multiples of 4.

Going up from 1 to 100 we get the following pairs of multiples of 4 followed by multiples of 9 and multiples of 9 followed by multiples of 4.

8, 9
27, 28
44, 45
63, 64
80, 81
99, 100

So we get 6 more.

4 + 6 = 10 total possible integers x such that x(x + 1) is a multiple of 36.

So actually the answer is (10 integers that work)/(100 possible integers) = 1/10.

LOL. I guess this question and explanation need some editing. Beyond that, while this question is pretty cool and good for practice, probably it takes a little more work than any similar one you might see on the actual test.
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by talaangoshtari » Tue Jun 16, 2015 12:37 pm
LCM(9, 12) = 36
when x is even, we have 2 numbers multiplied that 1 is odd and the other is even.
When x is odd, we have 2 numbers multiplied that 1 is odd and the other is even.
Since an even number has one 2, we actually have to find the probability of divisibility of the product by 18.
# of numbers divisible by 18 = [(90 - 18)/18] + 1 = 5
=> the probability is equal to 5/100
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by GMATGuruNY » Tue Jun 16, 2015 1:44 pm
talaangoshtari wrote:LCM(9, 12) = 36
when x is even, we have 2 numbers multiplied that 1 is odd and the other is even.
When x is odd, we have 2 numbers multiplied that 1 is odd and the other is even.
Since an even number has one 2, we actually have to find the probability of divisibility of the product by 18.
# of numbers divisible by 18 = [(90 - 18)/18] + 1 = 5
=> the probability is equal to 5/100
This line of reasoning isn't quite right.
To determine the number of options for x(x+1), we cannot simply count the multiples of 18 between 1 and 100.

Here are the multiples of 18 between 1 and 100:
18, 36, 54, 72, 90.
Here are the options for x(x+1) that are yielded by 18, 54 and 90:
17*18
18*19
53*54
54*55
89*90
90*91
None of these options is divisible by 12.
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by Matt@VeritasPrep » Tue Jun 23, 2015 7:05 pm
We're looking for the probability that x² + x is a multiple of 36.

That gives us x² + x = 36k, or x² + x - 36k = 0.

Since our middle term is x, our roots must differ by 1. So if one root is r, the other root is (r - 1).

The roots are consecutive integers, so they must be coprime. That means we have two cases:

1:: either r or (r - 1) is a multiple of 36

2:: r is a multiple of 4 and (r - 1) is a multiple of 9, or vice versa

The first case is easy: r = 36, 37, 72, 73. That gives sets of {35, 36}, {36, 37}, {71, 72}, and {72, 73}.

The second case is a little tiresome, but still workable. Consider all (r - 1) divisible by 9, which gives r = 10, 19, 28, 37, 46, 55, 64, 73, 82, 91, 100. Then eliminate all cases in which r is NOT divisible by 4, which leaves r = 28, 64, 100. Thus our sets are {27, 28}, {63, 64}, and {99, 100}.

We also have the other way round, in which r is a multiple of 9 and (r - 1) is divisible by 4. The first pass gives r = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, but the (r - 1) case will only hold for r = 9, 45, and 81. That gives sets of {8, 9}, {44, 45}, and {80, 81}.
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by Matt@VeritasPrep » Tue Jun 23, 2015 7:10 pm
Another way to do this would be to consider a little modular arithmetic. (This goes a little beyond the GMAT, but it works.)

To have x² + x = 0 mod 36, we'd have x in one of these forms, in which k is an arbitrary integer:

x = 36k
x = 36k + 8
x = 36k + 27
x = 36k + 35

From there, we'd take each multiple of 36 in range and count it up to four times, depending on how many of the terms are in our set. So for our purposes here, our multiples of 36 are 0, 36, and 72. For 0, we have {8, 27, 35}; for 36, we have {36, 44, 63, 71}; for 72, we have {72, 80, 99}.
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