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If a sequence of positive odd integers has six terms containing

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Director
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G
Joined: 07 Dec 2014
Posts: 815

Kudos [?]: 248 [0], given: 12

If a sequence of positive odd integers has six terms containing [#permalink]

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New post 04 Oct 2017, 19:15
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  45% (medium)

Question Stats:

60% (01:11) correct 40% (01:30) wrong based on 25 sessions

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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
[Reveal] Spoiler: OA

Kudos [?]: 248 [0], given: 12

Expert Post
Math Forum Moderator
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P
Joined: 02 Aug 2009
Posts: 4986

Kudos [?]: 5506 [0], given: 112

Re: If a sequence of positive odd integers has six terms containing [#permalink]

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New post 04 Oct 2017, 20:20
gracie 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



Hi...

6 consecutive integers having 13 digits MEANS 5 are 2-digit and 1 is 3-digit integer.
So smallest odd 3-digit integer is 101...
Therefore the smallest of 6 consecutive integers is 101-2*(6-1)=101-10=91..
91 = 13*7

So B is the answer.
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5506 [0], given: 112

Intern
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Joined: 15 Oct 2016
Posts: 4

Kudos [?]: [0], given: 1

Re: If a sequence of positive odd integers has six terms containing [#permalink]

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New post 04 Oct 2017, 20:40
Answer is B, i.e 13
Let us see how

Six odd terms making total of 13 digits, this means 6 term should be a 3 digit term and all the other terms to be less than 100
Arithmetic Progression of n terms:
S(nth)= a+(n-1)d
S(nth)= a+5d
Since terms of sequence are odd 5d should be even
Therefore d value will be 2
(Considering minimum value of first term be 11, if d=4 then 5th term also becomes 3 digit number, making total digits 14 )
Therefore S(nth) has to be 101 because it’s minimum 3 digit number of which 2 is subtracted gives a two digit number I.e 99
101=a+10
a=91
Which is a multiple of 13
13x7=91
Therefore B is the correct answer.
Please let me know if there is a shorter way to answer this question


Sent from my iPhone using GMAT Club Forum

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Intern
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B
Joined: 11 Aug 2015
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Location: India
Concentration: Operations, Statistics
GMAT 1: 560 Q38 V28
GMAT 2: 610 Q46 V28
Re: If a sequence of positive odd integers has six terms containing [#permalink]

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New post 05 Oct 2017, 01:26
This question is missing something..we can also have values as below.
11 29 47 65 83 101
13 31 49 67 85 103
15 33 51 69 87 105
17 35 53 71 89 107
19 37 55 73 91 109

Can some expert provide solution for this question.

Kudos [?]: 2 [0], given: 4

Manager
Manager
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Joined: 20 Jan 2016
Posts: 197

Kudos [?]: 12 [0], given: 63

Re: If a sequence of positive odd integers has six terms containing [#permalink]

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New post 11 Oct 2017, 11:24
chetan2u wrote:
gracie 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



Hi...

6 consecutive integers having 13 digits MEANS 5 are 2-digit and 1 is 3-digit integer.
So smallest odd 3-digit integer is 101...
Therefore the smallest of 6 consecutive integers is 101-2*(6-1)=101-10=91..
91 = 13*7

So B is the answer.


Why are we assuming they are consecutive integers?

Kudos [?]: 12 [0], given: 63

Re: If a sequence of positive odd integers has six terms containing   [#permalink] 11 Oct 2017, 11:24
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