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# In a sequence of N consecutive positive integers, are there exactly fo

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In a sequence of N consecutive positive integers, are there exactly fo  [#permalink]

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27 Jun 2015, 05:34
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In a sequence of N consecutive positive integers, are there exactly four multiples of 5?

(1) N=16
(2) The greatest number in the sequence is a multiple of 5.

How to solve this without plug respectively trial & error approach?

Thanks

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Re: In a sequence of N consecutive positive integers, are there exactly fo  [#permalink]

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27 Jun 2015, 18:22
Every fifth integer is a multiple of 5, so we want to know how many groups of five integers we have.

1) INSUFFICIENT. This seems to tell us that we have three groups of five (so three multiples of 5), but there's a remainder of 1, so we need to know where the sequence starts. If it starts on a multiple of 5, that last number at the end will also be a multiple of 5 (e.g. 5, 10, 15, 20). If it starts with a non-multiple of 5, then the remaining number will also be a non-multiple of 5 (e.g. 1 . . . 16 has only three multiples of 5: 5, 10, 15).

2) INSUFFICIENT. We have no idea how long the sequence is.

1&2) SUFFICIENT. If we have sixteen consecutive integers ending in a multiple of 5, then the first integer will also be a multiple of 5. Therefore, as mentioned above, there will be exactly four multiples of 5. (It could be the #s 5-20, 10-25, 15-30, etc.)
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In a sequence of N consecutive positive integers, are there exactly fo  [#permalink]

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27 Jun 2015, 21:31
reto wrote:
In a sequence of N consecutive positive integers, are there exactly four multiples of 5?

(1) N=16
(2) The greatest number in the sequence is a multiple of 5.

How to solve this without plug respectively trial & error approach?

Thanks

since the Q asks us four multipes of 5..
min consecutive numbers required is if the first is a multiple of 5.. another 3 multipes require 3*5=15 numbers..
total 16..

max consecutive numbers is if the first int gives a remainder of 1 when divided by 5.. so numbers required=5*4=20..

lets look at the statements..
1) stat 1 tells us the number of consecutive integer is 16.. so ans can be 'yes' if the first is a multiple of 5, otherwise 'no'... insuff
2) stat 1 tells us that the gretaest number is multiple of 5.. so ans can be 'yes' if the number of int is 16 or more , otherwise 'no'... insuff
combined.. we know that the number of consecutive integer is 16 and that the gretaest number is multiple of 5... so ans is 'yes'... suff
ans C
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In a sequence of N consecutive positive integers, are there exactly fo  [#permalink]

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16 Jul 2016, 02:36
reto wrote:
In a sequence of N consecutive positive integers, are there exactly four multiples of 5?

(1) N=16
(2) The greatest number in the sequence is a multiple of 5.

How to solve this without plug respectively trial & error approach?

Thanks

(1) N=16
Means there are total of 16 consecutive numbers in the set
If the set starts with 5 then S={5,6..................20,21} There will be 4 multiples of 5 namely 5,10,15,20 ==> yes
If the Set starts with 8 then S={8,9..................23,24} There will be 3 multiples of 5 namely 10,15,20 ==> no
INSUFFICIENT

(2) The greatest number in the sequence is a multiple of 5.
Meaning the last element is a multiple of 5
S={..............40} The rest of the elements will be {24,25,...............40} There will always be 4 multiples of 5 namely (25,30,35,40) if the last element is a multiple of 5.
SUFFICIENT

NOTE THAT :- If the first element is a multiple of 5 then also we will have a total of 4 multiples of 5
WHY :- Because the first or last term is a multiple and the remaining 15 consecutive numbers will atleast have 15/5 = 3 more multiples of 5. Hence FIRST/LAST+3=4
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Re: In a sequence of N consecutive positive integers, are there exactly fo  [#permalink]

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30 Jul 2017, 04:03
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Re: In a sequence of N consecutive positive integers, are there exactly fo   [#permalink] 30 Jul 2017, 04:03
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