DS

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 46
Joined: Thu Nov 22, 2007 6:32 am

DS

by ukr.net » Thu Dec 06, 2007 9:52 am
Is the sum of a series of n consecutive integers even?

1) n=6
2) The n digit number formed by the series is a multiple of nine.


ANS is D
What exactly does the tatement 2 means?
Source: — Data Sufficiency |

Master | Next Rank: 500 Posts
Posts: 214
Joined: Wed Nov 14, 2007 6:30 am
Thanked: 15 times

by sujaysolanki » Thu Dec 06, 2007 5:59 pm
For simplicity take n = 6 ..

Sum of first 6 consecutive terms = 6x +15

Now say x = 2 or x = 5 each time the sum is a multiple of 9 but the sum os not even Hence we get no ..hence it is sufficient

Master | Next Rank: 500 Posts
Posts: 128
Joined: Sat Nov 17, 2007 11:59 am
Thanked: 1 times

by cris » Sun Jan 06, 2008 5:24 pm
I still dont get it...what does x stands for?

Junior | Next Rank: 30 Posts
Posts: 24
Joined: Sun Jan 07, 2007 6:14 pm
Thanked: 1 times

by Argen » Mon Jan 07, 2008 7:26 am
1) For this statement we need to know that an odd number + another odd number = an even number, and an odd number + an even number = an odd number. If n = 6, that means there are 3 pairs of an odd and an even number. When we sum up each of the three pairs, we have 3 odd numbers. When we sum up the 3 odd numbers, we have an odd number.

2) I think this statement means if we take all the numbers and use them as digits to form 1 number, then that number is a multiple of 9. For example, 9, 45, 234, etc, fit the criteria. We also know for a number to be a multiple of 9, the sum of its digits must equal 9, which is an odd number.

Therefore the answer is D.
Last edited by Argen on Mon Jan 07, 2008 12:42 pm, edited 1 time in total.

Master | Next Rank: 500 Posts
Posts: 128
Joined: Sat Nov 17, 2007 11:59 am
Thanked: 1 times

by cris » Mon Jan 07, 2008 10:22 am
Thanks a lot Argen! I did not understand what statement 2 could possibly mean....