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Re: For each positive integer k, let Sk denote the increasing arithmetic [#permalink]
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Bunuel wrote:
For each positive integer k, let Sk denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is k. For example, S3 is the sequence 1, 4, 7, 10 ….. What is the number of values of k for which Sk contain the term 36 ?

A. 6
B. 8
C. 12
D. 16
E. 24


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sthahvi

The options do not contain the correct answer.

Now k is the common difference between consecutive terms.
Whatever be the difference between 1 and 36 should be equally distributable in k parts, then only k would be an integer.
36-1=35, so factors are 1,5,7 and 35.
The sequences are
1,2,3,4...35,36,37...
1,6,11,16....31,36,41...
1,8,15,22,29,36,43,...
1,36,71...

Since no options are anywhere close to 4, 36 must be some other value.

If I look at the options, 361 is one such value.
\(361-1=360=2^3*3^2*5\)
Number of factors = (3+1)(2+1)(1+1)=24

E
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Re: For each positive integer k, let Sk denote the increasing arithmetic [#permalink]
The tough part is understanding the question layout and what the defined sequence really is.

The only answer it could possibly be is (E) because the max value for each Sk listed other than 24 is too low and less than 361.

(D)16

For S15, you will get 16 values.

The 1st value will be 1

Then the 2nd value will be: 1 + (1)(15) = 16

The 3rd value will be: 1 + (2)(15) = 31

And the 16th value will be: 1 + (15)(15) = 226

So the max number in the sequence is too low.

The answer must be E.

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Re: For each positive integer k, let Sk denote the increasing arithmetic [#permalink]
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Re: For each positive integer k, let Sk denote the increasing arithmetic [#permalink]
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