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If c is randomly chosen from the integers 20 to 99, inclusi
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26 Aug 2017, 02:19
26 Aug 2017, 02:19
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Question Stats:
24% (02:20) correct
75% (03:48) wrong based on 65 sessions
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If c is randomly chosen from the integers 20 to 99, inclusive, what is the probability that \((c^3-c)\) is divisible by 12?
enter your value as a fraction
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\(\frac{3}{4}\)
Re: If c is randomly chosen from the integers 20 to 99, inclusi
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31 Aug 2017, 08:36
31 Aug 2017, 08:36
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This is a really tough question.
Starting from what we do have
\(\frac{favorable outcomes}{total # of possibilities}\)
99-20+1=80 possibilities
Now a number of being divisible by 12 must contain in it 2*2*3.
Here is probably the most difficult thing to realize \(c^3 - c\) is the same of having \(c(c^2 - 1)\) = \((c-1)c(c+1)\)
This latter expression means you do have a set of 3 consecutive integers divisible by 12. A rule is: every 3 consecutive integers must be divisible by 3
Actually, the question really boils down: how many numbers you pick from are divisible by \(2*2=4\).
In this case, c can be 21,23,25.............95,97,99 so \(\frac{99-21}{2}\) + 1 = 40
OR 20,24,26...........92,96 so \(\frac{96-20}{4}\) + 1 = 20
Which means \(\frac{40+20}{80}\) = \(\frac{60}{80}\) = \(\frac{3}{4}\)
Ask if something is unclear.
Regards
Starting from what we do have
\(\frac{favorable outcomes}{total # of possibilities}\)
99-20+1=80 possibilities
Now a number of being divisible by 12 must contain in it 2*2*3.
Here is probably the most difficult thing to realize \(c^3 - c\) is the same of having \(c(c^2 - 1)\) = \((c-1)c(c+1)\)
This latter expression means you do have a set of 3 consecutive integers divisible by 12. A rule is: every 3 consecutive integers must be divisible by 3
Actually, the question really boils down: how many numbers you pick from are divisible by \(2*2=4\).
In this case, c can be 21,23,25.............95,97,99 so \(\frac{99-21}{2}\) + 1 = 40
OR 20,24,26...........92,96 so \(\frac{96-20}{4}\) + 1 = 20
Which means \(\frac{40+20}{80}\) = \(\frac{60}{80}\) = \(\frac{3}{4}\)
Ask if something is unclear.
Regards
Re: If c is randomly chosen from the integers 20 to 99, inclusi
[#permalink]
20 Jun 2021, 08:31
20 Jun 2021, 08:31
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There are 99 – 20 + 1 = 80 possible values for c, so the unknown is how
many of these c values yield a c3 – c that is divisible by 12.
The prime factorization of 12 is 2 × 2 × 3. There are several ways of thinking
about this: numbers are divisible by 12 if they are divisible by 3 and by 2
twice, or if they are multiples of both 4 and 3, or if half of the number is an
even multiple of 3, etc.
The expression involving c can be factored:
c3 – c = c(c2 – 1) = c(c – 1)(c + 1)
These are consecutive integers. It may help to put them in increasing order: (c– 1)c(c + 1). Thus, this question has a lot to do with consecutive integers, and not only because the integers 20 to 99 themselves are consecutive.
In any set of three consecutive integers, a multiple of 3 will be included.
Thus, (c – 1)c(c + 1) is always divisible by 3 for any integer c. This takes care
of part of the 12. So the question becomes “How many of the possible (c –
1)c(c + 1) values are divisible by 4?” Since the prime factors of 4 are 2’s, it
makes sense to think in terms of odds and evens.
(c – 1)c(c + 1) could be (E)(O)(E), which is definitely divisible by 4, because
the two evens would each provide at least one separate factor of 2. Thus, c3 –c is divisible by 12 whenever c is odd, which are the cases c = 21, 23,25, …,95, 97, 99. That’s 40 possibilities.
Alternatively, (c – 1)c(c + 1) could be (O)(E)(O), which will only be divisible
by 4 when the even term itself is a multiple of 4. Thus, c3 – c is also divisible
by 12 whenever c is a multiple of 4, which are the cases c = 20, 24, 28, …,92, 96. That’s 20 possibilities.
The probability 40+20/80 is thus 3/4.
many of these c values yield a c3 – c that is divisible by 12.
The prime factorization of 12 is 2 × 2 × 3. There are several ways of thinking
about this: numbers are divisible by 12 if they are divisible by 3 and by 2
twice, or if they are multiples of both 4 and 3, or if half of the number is an
even multiple of 3, etc.
The expression involving c can be factored:
c3 – c = c(c2 – 1) = c(c – 1)(c + 1)
These are consecutive integers. It may help to put them in increasing order: (c– 1)c(c + 1). Thus, this question has a lot to do with consecutive integers, and not only because the integers 20 to 99 themselves are consecutive.
In any set of three consecutive integers, a multiple of 3 will be included.
Thus, (c – 1)c(c + 1) is always divisible by 3 for any integer c. This takes care
of part of the 12. So the question becomes “How many of the possible (c –
1)c(c + 1) values are divisible by 4?” Since the prime factors of 4 are 2’s, it
makes sense to think in terms of odds and evens.
(c – 1)c(c + 1) could be (E)(O)(E), which is definitely divisible by 4, because
the two evens would each provide at least one separate factor of 2. Thus, c3 –c is divisible by 12 whenever c is odd, which are the cases c = 21, 23,25, …,95, 97, 99. That’s 40 possibilities.
Alternatively, (c – 1)c(c + 1) could be (O)(E)(O), which will only be divisible
by 4 when the even term itself is a multiple of 4. Thus, c3 – c is also divisible
by 12 whenever c is a multiple of 4, which are the cases c = 20, 24, 28, …,92, 96. That’s 20 possibilities.
The probability 40+20/80 is thus 3/4.
