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Is the sum of a series of six consecutive integers even? (1) [#permalink]
02 Dec 2006, 11:31
00:00
A
B
C
D
E
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
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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.
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
S = [2a + (n-1)d]/2 * n here a is the first term, n = 6, d = 1 S = (2a + 5)3For any value of a, S is always odd
I donâ€™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
S = [2a + (n-1)d]/2 * n here a is the first term, n = 6, d = 1 S = (2a + 5)3For any value of a, S is always odd
I donâ€™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
Take a as the first term
1st - a
2nd â€“ a + d
3rd â€“ a + 2d
...
6th â€“ a + 5d
..
nth â€“ 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
Take a as the first term 1st - a 2nd â€“ a + d 3rd â€“ a + 2d ... 6th â€“ a + 5d .. nth â€“ 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