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Is the sum of a series of six consecutive integers even? (1)

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Is the sum of a series of six consecutive integers even? (1) [#permalink] New post 02 Dec 2006, 11:31
00:00
A
B
C
D
E

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Is the sum of a series of six consecutive integers even?

(1) The first integer in the series is even
(2) The six digit number formed by the series is a multiple of nine

Answer to this one is apparently D, but I'm getting A. Can you guys check my logic for errors?

(1) pretty easy to determine this is sufficient
(2) using properties of numbers, numbers that are divisible by nine have the property that the sum of the digits that make up that number are also divisible by 9 (e.g., 27, 2 + 7 = 9). With that said, the sum of the digits of the six-digit number formed by the series must be divisible by 9. Numbers divisible by 9 can be even or odd, so this is insufficient.
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Re: DS [#permalink] New post 02 Dec 2006, 12:13
artshep wrote:
Is the sum of a series of six consecutive integers even?

(1) The first integer in the series is even
(2) The six digit number formed by the series is a multiple of nine

Answer to this one is apparently D, but I'm getting A. Can you guys check my logic for errors?

(1) pretty easy to determine this is sufficient
(2) using properties of numbers, numbers that are divisible by nine have the property that the sum of the digits that make up that number are also divisible by 9 (e.g., 27, 2 + 7 = 9). With that said, the sum of the digits of the six-digit number formed by the series must be divisible by 9. Numbers divisible by 9 can be even or odd, so this is insufficient.


1) If first int even then sequence is even,odd,even,odd,even,odd and so for any 6 consecutive starting with an even no. the total is odd. SUF

2) I found it easier to pick numbers:
Sum of 1-6 = 21 Not div by 9
Sum of 2-7 = 27 Divisible by nine and odd
Sum of 3-8 = 33 Not div by 9
Sum of 4-9 = 39 Not div by 9
Sum of 5-10 = 45 Divisible by nine and odd
The series continues going up by 6 with each multiple of 9 being odd so SUFF
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 [#permalink] New post 02 Dec 2006, 12:18
This is a strange problem..

S = [2a + (n-1)d]/2 * n here a is the first term, n = 6, d = 1
S = (2a + 5)3
For any value of a, S is always odd

I don’t need any extra hint to answer this..where I am going wrong?

Last edited by anindyat on 02 Dec 2006, 12:24, edited 1 time in total.
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 [#permalink] New post 02 Dec 2006, 12:23
anindyat wrote:
This is a strange problem..

S = [2a + (n-1)d]/2 * n here a is the first term, n = 6, d = 1
S = (2a + 5)3
For any value of a, S is always odd

I don’t need any extra hint to answer this..where am I going wrong?


Where did you get that formula and could you explain it a little more?

Thanks!
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 [#permalink] New post 02 Dec 2006, 12:30
MBAlad wrote:
anindyat wrote:
This is a strange problem..

S = [2a + (n-1)d]/2 * n here a is the first term, n = 6, d = 1
S = (2a + 5)3
For any value of a, S is always odd

I don’t need any extra hint to answer this..where am I going wrong?


Where did you get that formula and could you explain it a little more?

Thanks!


This is the sum of AP series, I am sure you also know this...
S = (first term + last teram) / 2 * number of terms
Here first teram a
last term = a + (n-1)d where d is the increment
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 [#permalink] New post 02 Dec 2006, 12:49
anindyat wrote:
MBAlad wrote:
anindyat wrote:
This is a strange problem..

S = [2a + (n-1)d]/2 * n here a is the first term, n = 6, d = 1
S = (2a + 5)3
For any value of a, S is always odd

I don’t need any extra hint to answer this..where am I going wrong?


Where did you get that formula and could you explain it a little more?

Thanks!


This is the sum of AP series, I am sure you also know this...
S = (first term + last teram) / 2 * number of terms
Here first teram a
last term = a + (n-1)d where d is the increment


the formula is derived for terms starting with n=0 for the first and so on..
for 6 numbers n must be 5
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 [#permalink] New post 02 Dec 2006, 12:55
Quote:
the formula is derived for terms starting with n=0 for the first and so on..
for 6 numbers n must be 5


This is an area I have covered yet. Why for 6 numbers, n = 5?
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 [#permalink] New post 02 Dec 2006, 13:15
MBAlad wrote:
Quote:
the formula is derived for terms starting with n=0 for the first and so on..
for 6 numbers n must be 5


This is an area I have covered yet. Why for 6 numbers, n = 5?


You have 6 terms 0th,1st,2nd,3rd,4th,5th
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 [#permalink] New post 02 Dec 2006, 13:19
Anindyat,

I'm afraid I'm not understanding your formula. Can you go into detail on what this is used for, how it's used, and how you applied it to this problem?
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 [#permalink] New post 02 Dec 2006, 21:45
Take a as the first term
1st - a
2nd – a + d
3rd – a + 2d
...
6th – a + 5d
..
nth – a + (n-1)d

Sum = (first term + last term) / 2 * n
=> [a + a + (n-1)d]/2 * n
For consecutive integers, d =1
Sum = [2a + (n-1)]/2 * n
For 6 consecutive integers Sum = ([2a + (n-1)]/2) * n = ((2a + 5)/2)*6= (2a + 5)3
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 [#permalink] New post 02 Dec 2006, 23:50
seriously, do we even need the conditions?

The question itself is Suff.
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 [#permalink] New post 03 Dec 2006, 00:24
sum of any series that involves addtion can be found using the following formula

SUM= n*(First term+last term)/2
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 [#permalink] New post 03 Dec 2006, 04:46
Six consecutive integers have 3 odd and 3 even integers. No matter what the nos. are the total is always odd. We don't even need the S1&2
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 [#permalink] New post 03 Dec 2006, 07:39
anindyat wrote:
Take a as the first term
1st - a
2nd – a + d
3rd – a + 2d
...
6th – a + 5d
..
nth – a + (n-1)d

Sum = (first term + last term) / 2 * n
=> [a + a + (n-1)d]/2 * n
For consecutive integers, d =1
Sum = [2a + (n-1)]/2 * n
For 6 consecutive integers Sum = ([2a + (n-1)]/2) * n = ((2a + 5)/2)*6= (2a + 5)3



Thanks anindyat - that was really helpful
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 [#permalink] New post 03 Dec 2006, 08:20
How would you calculate the product of an arithmetic progression?
  [#permalink] 03 Dec 2006, 08:20
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