Anurag@Gurome wrote:

rnaah wrote:If c is randomly chosen from the integers 20 to 99, inclusive, what is the probability that c^3 - c is divisible by 12?

Total number of integers in the range = (99 - 20) + 1 = 80

c³ - c = (c - 1)c(c + 1) = Product of three consecutive integers.
Now, product of any three consecutive integers is always divisible by 3.

Now, if c is of the form 4n, (4n + 1) or (4n + 3), then (c³ - c) will be divisible by 4 also and hence by 12.

But if c is of the corm (4n + 2), then (c³ - c) = (c - 1)c(c + 1) = (4n + 1)(4n + 2)(4n + 3) will be divisible by 2 not by 4. Hence, (c³ - c) will be divisible by 6 not by 12.

Number of integers of the form (4n + 2) between 20 and 99 inclusive = smallest integer greater than (99 - 20)/4 = smallest integer greater than 19.(something) = 20

Hence, for (80 - 20) = 60 integers (c³ - c) will be divisible by 12.

Hence, required probability = 60/80 = 3/4

I do not understand from the bold text downwards. could u explain in simple language?

While I see that the smallest integer >=20 of the form 4n+2 is 22 and the largest integer of the same form =<99 is 98. This gives a range of n from 5 to 24, and the total to 24-5+1 = 20.

Then the probability can be calculated in the same way as your explanation.

Number of integers of the form (4n + 2) between 20 and 99 inclusive = smallest integer greater than (99 - 20)/4 = smallest integer greater than 19.(something) = 20

The quoted part is what I did not understand. Is this a general method to calculate the total number of integers in an evenly spaced set? (Last Element - First Element)/Set Distance = Total? Yeah, come to think of it, it now makes sense!!

Well, thank you!

any 3 consecutive integer always divisible by 3...
so if 3 numbers are like Even*odd*even then it will always be divisible by 12.
i.e. if c is odd then c-1 and c+1 will be even and hence multiplicationn will be divisuble by 12.

since there are 40 odd numbers.
probality is 40/80 = 1/2

Please help where am i wrong ?

Thanks

c^3-c=(c-1)c(c+1) Note that this is the product of three consecutive integers.

If the product is divisible by 12, it must be divisible by 3 and 4.

The product of any three consecutive integers is divisible by 3, so we can really think of the problem as being reduced to "is C^3-1 divisible by 4?"

First, note that 20-99 starts with an even number and ends with an odd number, so there must be an equal number of evens and odds. The even numbers (20,22,24...98) starts with a number that is divisible by 4 and ends with a number that is not divisible by 4, so there must be an equal number of each.

Let's consider how the product could FAIL to be divisible by 4. If c is odd, you would have even*odd*even which is always divisible by 4, so for it to fail, c must be one of the even numbers that is NOT divisible by 4.

1/2 the numbers are even, and 1/2 of those are not divisible by 4, so 1/4 of the possible c values from 20-99 would make c^3-1 fail to be divisible by 12.

1-1/4=3/4, so there is a 3/4 chance of c^3-1 being divisible by 12.

1947 wrote:any 3 consecutive integer always divisible by 3...
so if 3 numbers are like Even*odd*even then it will always be divisible by 12.
i.e. if c is odd then c-1 and c+1 will be even and hence multiplicationn will be divisuble by 12.

since there are 40 odd numbers.
probality is 40/80 = 1/2

Please help where am i wrong ?

Thanks

You forgot to account for the even values of c, like 20, 24, 28.... that make c^3-1 divisible by 12.