c³-c
(c-1)c(c+1)
This can be said to be products of 3 consecutive numbers
Product of 3 consecutive integers is always divisible by 6.
For product not to be divisible by 12
we need consecutive triplets such as these
21,22,23
25,26,27
29,30,31
33,34,35
.
.
.
97,98,99
The first and third numbers should be odd
the second number should not be divisible by 4
all other possible triplets will have product whic h is divisible by 12
acc to the conditions, possible values of c can be 22,26,30,....98
For this A.P.,
a=22
l=98
d=4
l = a+(n-1)d
98 = 22+(n-1)4
76/4 = n-1
19 = n-1
n=20
The probability that the product is not divisible by 12 = 20/80 = 1/4
The pprobability that c³-c is divisible by 12 = 1 - 1/4 = [spoiler]3/4[/spoiler]
Probability
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pemdas
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simplify such as c(c-1)(c+1)/2*2*3; (c-1)(c+1) is always divisible by 3 and we need to account for divisibility of either c(c-1) or c(c+1) by 4. If we start listing samples 5,6,7,8 then 3 numbers out of 4 will be divisible by 4 always this makes 3/4 throughout the number line divisibility for either c(c-1) or c(c+1). The probability will be the same, 3/4
knight247 wrote:If c is a randomly chosen from the integers 20 to 99, inclusive, what is the probability that
c³-c is divisible by 12?
Answer is [spoiler]3/4[/spoiler]
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To be divisible by 12, c³-c must be a multiple of 3 and 4.knight247 wrote:If c is a randomly chosen from the integers 20 to 99, inclusive, what is the probability that
c³-c is divisible by 12?
Answer is [spoiler]3/4[/spoiler]
c³-c = c(c²-1) = c(c-1)(c+1).
c-1, c, and c+1 are three consecutive integers.
Of every three consecutive integers, exactly one is a multiple of 3.
Thus, (c-1)(c)(c+1) is a multiple of 3.
The question is whether (c-1)(c)(c+1) will be a multiple of 4.
If c is odd, then (c-1)(c)(c+1) = (even)(odd)(even).
In this case, the product will be a multiple of 4, since it includes 2 even factors.
If c is a multiple of 4, then (c-1)(c)(c+1) = (odd)(multiple of 4)(odd).
In this case, the product will be a multiple of 4, since one of its factors is a multiple of 4.
If c is an even integer that is NOT a multiple of 4, then (c-1)(c)(c+1) = (odd)(even non-multiple of 4)(odd).
In this case, (c-1)(c)(c+1) will NOT be a multiple of 4.
Between 20 and 99:
The total number of integers = 80.
Since half are even, the number of even integers = 40.
Since every other even integer is a multiple of 4, the number of even integers that are NOT multiples of 4 = 20.
Thus, P(even non-multiple of 4) = 20/80 = 1/4.
Since any other value for c will yield a multiple of 4 for (c-1)(c)(c+1):
P(c³-c will be a multiple of 12) = 1-1/4 = 3/4.
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