If a sequence of positive odd integers has six terms contain

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alanforde800Maximus: Master | Next Rank: 500 Posts; Posts: 187; Joined: Tue Sep 13, 2016 12:46 am

by alanforde800Maximus » Fri Jul 27, 2018 11:10 pm

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If a sequence of positive odd integers has six terms containing a total of thirteen digits, then the first term is a multiple of?

A. 11
B. 13
C. 15
D. 17
E. 19

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Jul 28, 2018 2:21 am

alanforde800Maximus wrote:If a sequence of positive odd integers has six terms containing a total of thirteen digits, then the first term is a multiple of?

A. 11
B. 13
C. 15
D. 17
E. 19

This problem is invalid.
Since the only conditions are that six terms must be positive and odd and have a total of 13 digits, an infinite number of cases are possible.
Here are two options:
11, 13, 15, 17, 19, 101
13, 11, 15, 17, 19, 101
Since the sequence does not have to be ascending, any of the answer choices could be a divisor of the first term.
Ignore this problem.
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Mathexpert: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Mon Jun 22, 2015 7:25 am

by Mathexpert » Thu Aug 16, 2018 7:06 am

I agree the problem is unsolvable by the way it's stated. However, I think he meant to say "positive consecutive odd integers" instead of "positive odd integers".

GMATGuruNY wrote:
alanforde800Maximus wrote:If a sequence of positive odd integers has six terms containing a total of thirteen digits, then the first term is a multiple of?

A. 11
B. 13
C. 15
D. 17
E. 19
This problem is invalid.
Since the only conditions are that six terms must be positive and odd and have a total of 13 digits, an infinite number of cases are possible.
Here are two options:
11, 13, 15, 17, 19, 101
13, 11, 15, 17, 19, 101
Since the sequence does not have to be ascending, any of the answer choices could be a divisor of the first term.
Ignore this problem.
What is the source?

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