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If integer x is chosen at random from 50 to 149 inclusive, w
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26 Oct 2020, 14:19
26 Oct 2020, 14:19
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If integer x is chosen at random from 50 to 149 inclusive, what is the probability that x³ - x is a multiple of 12?
Enter your answer as a fraction
Note: One of my students asked me to solve this question. I thought I'd post it here, so that others can try it as well.
Enter your answer as a fraction
Note: One of my students asked me to solve this question. I thought I'd post it here, so that others can try it as well.
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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If integer x is chosen at random from 50 to 149 inclusive, w
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26 Oct 2020, 14:51
26 Oct 2020, 14:51
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GreenlightTestPrep wrote:
If integer x is chosen at random from 50 to 149 inclusive, what is the probability that x³ - x is a multiple of 12?
Enter your answer as a fraction
Enter your answer as a fraction
Show: ::
3/4
The number of integers from x to y inclusive equals y - x + 1
So, number of integers from 50 to 149 inclusive = 149 - 50 + 1 = 100
12 = (2)(2)(3)
So, in order for (x - 1)(x)(x + 1) to be a multiple of 12, we need at least two 2's and one 3 "hiding" in the product (x - 1)(x)(x + 1)
Now notice that x³ - x = x(x² - 1) = x(x + 1)(x - 1) = (x - 1)(x)(x + 1)
Notice that x-1, x and x+1 are three consecutive integers
Nice rule: The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
This means (x - 1)(x)(x + 1) is divisible by 3 for ALL integer values of x.
Since (x - 1)(x)(x + 1) is guaranteed to be divisible by 3, we just need to find the probability that (x - 1)(x)(x + 1) is divisible by 4
Now notice that HALF of the products (x - 1)(x)(x + 1) are in the form (EVEN)(ODD)(EVEN) and half are in the form (ODD)(EVEN)(ODD)
Let's examine each case separately.
Case i: The product (x - 1)(x)(x + 1) is in the form (EVEN)(ODD)(EVEN)
We can write: (x - 1)(x)(x + 1) = (some multiple of 2)(ODD)(some multiple of 2) = (2)(2)(some integer) = (4)(some integer)
This means (x - 1)(x)(x + 1) is a multiple of 4 every time x - 1 is EVEN
Since ALL products in the form (EVEN)(ODD)(EVEN) are divisible by 4 (and by 3), we can conclude that these products are divisible by 12.
Since even and odd numbers alternate, we can see that HALF of 100 possible products will be such that x - 1 is EVEN
In other words 50 products are such that x - 1 is EVEN (which means those products are divisible by 12 too)
Case ii: The product (x - 1)(x)(x + 1) is in the form (ODD)(EVEN)(ODD)
Also notice that the product can also have the form (ODD)(EVEN)(ODD)
SOME of the products in this form are divisible by 4, and SOME of the products are not divisible by 4.
Products in the form (ODD)(EVEN)(ODD) will be divisible by 4 when the middle number is divisible by 4.
For example, (59)(60)(61) is divisible by 4, since 60 is divisible by 4.
Likewise, (61)(62)(63) is NOT divisible by 4, since 62 is NOT divisible by 4.
And (63)(64)(65) is divisible by 4, since 64 is divisible by 4.
And (67)(68)(69) is NOT divisible by 4, since 68 is NOT divisible by 4.
As you can see, HALF of the products in the form (ODD)(EVEN)(ODD) are divisible by 4, and half of the products the form (ODD)(EVEN)(ODD) are NOT divisible by 4
50 of the 100 products are in the form (ODD)(EVEN)(ODD)
And HALF of those 50 are divisible by 4.
So, 25 products in the form (ODD)(EVEN)(ODD) are divisible by 4 (which means those products are divisible by 12 too)
So the total number of products that are divisible by 12 = 50 + 25 = 75
The probability that (x - 1)(x)(x + 1) is divisible by 12 = 75/100 = 3/4
Cheers,
Brent
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Re: If integer x is chosen at random from 50 to 149 inclusive, w
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Updated on: 27 Oct 2020, 13:57
Updated on: 27 Oct 2020, 13:57
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Here, x³ - x = x(x² - 1) = x(x + 1)(x - 1) = (x - 1)(x)(x + 1)
= Product of 3 consecutive integers.
And Product of 3 consecutive integers are always divisible by 3! i.e. 6. But, the question asked about the divisibility of 12. If any of the numbers are divisible by 4, the product will be divisible by 12 aw well. Now the question becomes-Whether any of the numbers is divisible by 4 as well?
Now check some numbers: [Rule of thumbs: check 4 cases for these probability type questions. If still confused, check 8 cases ]
(a) 49*50*51= NOT divisible by 12, coz no one is divisible by 4 (Here x=50, so x-1=49, x+1=51)
(b) 50*51*52= YES. it is divisible by 12, coz 52 is divisible by 4.
(c) 51*52*53= YES. it is divisible by 12, coz 52 is divisible by 4.
(d) 52*53*54= YES. it is divisible by 12, coz 52 is divisible by 4.
Here we can see that out of every 4 cases of product of 3 consecutive integers, 3 cases are divisible by 4 (consequently by 12). So 75% probability. so Answer is 0.75
= Product of 3 consecutive integers.
And Product of 3 consecutive integers are always divisible by 3! i.e. 6. But, the question asked about the divisibility of 12. If any of the numbers are divisible by 4, the product will be divisible by 12 aw well. Now the question becomes-Whether any of the numbers is divisible by 4 as well?
Now check some numbers: [Rule of thumbs: check 4 cases for these probability type questions. If still confused, check 8 cases ]
(a) 49*50*51= NOT divisible by 12, coz no one is divisible by 4 (Here x=50, so x-1=49, x+1=51)
(b) 50*51*52= YES. it is divisible by 12, coz 52 is divisible by 4.
(c) 51*52*53= YES. it is divisible by 12, coz 52 is divisible by 4.
(d) 52*53*54= YES. it is divisible by 12, coz 52 is divisible by 4.
Here we can see that out of every 4 cases of product of 3 consecutive integers, 3 cases are divisible by 4 (consequently by 12). So 75% probability. so Answer is 0.75
Originally posted by BelalHossain on 27 Oct 2020, 13:48.
Last edited by BelalHossain on 27 Oct 2020, 13:57, edited 1 time in total.
Last edited by BelalHossain on 27 Oct 2020, 13:57, edited 1 time in total.
