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23 Jan 2015, 07:25
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
54% (02:56) correct
46%
(02:48)
wrong
based on 345
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A term an is called a “cusp” of a sequence if \(a_n\) is an integer but \(a_{n+1}\) is not an integer. If \(a_5\) is a cusp of the sequence a1,a2,…,an,… in which \(a_1=k\) and \(a_n=-2\frac{(a_{n-1})}{3}\) for all n>1, then k could be equal to:
A. 3
B. 16
C. 108
D. 162
E. 243
Kudos for a correct solution.
A. 3
B. 16
C. 108
D. 162
E. 243
Kudos for a correct solution.
23 Jan 2015, 08:13
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ans D
Now, it can be seen that each term gets multiplied by -2/3. We are concerned here with divisiblity by 3.
a5 is int, but a6 is not an int. So a5 should be div by 3 and a6 not.
If k is the starting term, the terms keeping just divisiblity by 3 in mind become.
k, k/3, k/3^2, k/3^3, k/3^4, k/3^5
Thus k/3^4 or k/81 is integer but k/243 is not.
In the options, 162 /81 =2, but 162/243 is not an integer.
Now, it can be seen that each term gets multiplied by -2/3. We are concerned here with divisiblity by 3.
a5 is int, but a6 is not an int. So a5 should be div by 3 and a6 not.
If k is the starting term, the terms keeping just divisiblity by 3 in mind become.
k, k/3, k/3^2, k/3^3, k/3^4, k/3^5
Thus k/3^4 or k/81 is integer but k/243 is not.
In the options, 162 /81 =2, but 162/243 is not an integer.
26 Jan 2015, 04:21
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Bunuel
VERITAS PREP OFFICIAL SOLUTION:
As with many sequence problems, the wording of this question is designed to be horribly abstract. Let’s just focus on the fact that a5 is an integer but a6 is not an integer.
Putting the sequence into words can help. What do we do to find the next value? We take the previous value and multiply by −2/3. Or even more colloquially, we:
First, flip the sign of the previous one. Second, multiply it by two. Third, divide it by three.
So the second term will be −2/3 times the first one. The third term will be −2/3 times the second one. And so on.
Now, keep in mind that our goal is to find a value of k that will be an integer by the fifth term but not an integer by the sixth term. And this can make your job a lot easier. Positive vs. negative won’t have any bearing on whether a term is an integer or not, and the major operation that will control that will be division. Consider k = 6 as an example (since there are 2s and 3s all over this problem let’s pick a number that’s divisible by both). The first term would be 6, then the second term would be −2/3(6) or -4. Then the third term won’t be an integer, as the “divide by 3” part of the operation will leave you with a decimal.
So when you see that the 5th term of the sequence is going to be:
(−2/3)(−2/3)(−2/3)(−2/3)k, and will be an integer, but that the 6th term of the sequence will be:
(−2/3)(−2/3)(−2/3)(−2/3)(−2/3)k, and will not be an integer, you should see that the real question is “how many times can we evenly divide k by 3?”.
The answer, then, is 4, as it’s that fifth iteration of −2/3 that takes us from “integer” to “noninteger” status.
So kmust contain exactly four 3s, since 81=3^4 and since a5 contains no 3s (recall that a5 was an integer but not a multiple of 3).
Now, factor the answer choices. 3 is already prime factored, but it’s far short of four 3s, so it’s incorrect. 16=2^4; this has no 3s at all, so it’s also incorrect. 108=2^2*3^3; this answer is closer, but it’s still one 3 short, so it’s incorrect too. 162=(2)3^4; that’s the four 3s we need, so this is the correct answer. Finally, 243=3^5; that’s one 3 too many – we needed exactly four – so this is also incorrect.
Answer: D.
