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
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions sthahviThe 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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