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If a sequence of positive odd integers has six terms containing

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gracie
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If a sequence of positive odd integers has six terms containing [#permalink] New post   04 Oct 2017, 19:15
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If a sequence of positive odd consecutive 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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B
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chetan2u
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Re: If a sequence of positive odd integers has six terms containing [#permalink] New post   04 Oct 2017, 20:20
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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.
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Re: If a sequence of positive odd integers has six terms containing [#permalink] 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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Re: If a sequence of positive odd integers has six terms containing [#permalink] 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.
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Re: If a sequence of positive odd integers has six terms containing [#permalink] 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?
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stne
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Re: If a sequence of positive odd integers has six terms containing [#permalink] New post   14 Feb 2018, 08:31
pra1785 wrote:
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?


Absolutely I concur ! why are we assuming they are consecutive ?

6 odd terms , yielding 13 digits could also be 11, 15, 19, 35, 37, 971, but I guess in this case the question could not be solvable.

This question could simply be rewritten as :

Find the multiple of the smallest term of the 6 consecutive odd terms , which in total have 13 digits .
Hope this helps!
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Re: If a sequence of positive odd integers has six terms containing [#permalink] New post   18 Dec 2019, 02:41
Since it is a sequence of six positive odd consecutive integers and the number six terms contain a total of 13 digits, we can assume that 5 out of the 6 numbers will have 2 digits and one will have 3 digits.

The only secuence that satisfies this (because its odd consecutive integers) will start with the lowest 3 digit odd number, 101. Therefore we need to add the 5 previous consecutive odd numbers to the secuence to find the first number of the secuence (or we can just do 101 -2*5). The secuence will be 91, 93, 95, 97, 99, 101.

91 is a multiple of 13 therefore the answer is B

So the first number is 91 which
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Re: If a sequence of positive odd integers has six terms containing [#permalink] New post   07 Apr 2021, 19:25
If a sequence of positive odd consecutive 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

Start by taking the extreme scenario. Suppose all 6 terms have 2 digits each.

6 x 2 = 12 <--- We are short one. This means the last term must have three digits

91, 93, 95, 97, 99, 101

91 / 13 = k <-- k is an integer

B.
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Re: If a sequence of positive odd integers has six terms containing [#permalink]
07 Apr 2021, 19:25
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