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14,15,16 =45 + already we have n=2 and n=6 So insufficient

Together

we have atleast 2 option where n is even and less than 9

so E

thank you for the expanation. Do you know any shortcut for picking numbers? it took me some time to check on different combinations and we cannot afford it during the test. thanks!

Re: CONSECUTIVE INTEGERS.. [#permalink]
09 Aug 2009, 21:54

2

This post received KUDOS

We can pick numbers as follows

Case 1) divide the number by 2 and play around with the numbers around it to get the sum Case 2) divide the number by 3 and play around with the numbers around it to get the sum Case 3) divide the number by 4 and play around with the numbers around it to get the sum Case 4) divide the number by 5 and play around with the numbers around it to get the sum and go on ... till you realize that sum of the n numbers exceeds 45

In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46

Re: CONSECUTIVE INTEGERS.. [#permalink]
09 Aug 2009, 22:29

gmanjesh wrote:

We can pick numbers as follows

Case 1) divide the number by 2 and play around with the numbers around it to get the sum Case 2) divide the number by 3 and play around with the numbers around it to get the sum Case 3) divide the number by 4 and play around with the numbers around it to get the sum Case 4) divide the number by 5 and play around with the numbers around it to get the sum and go on ... till you realize that sum of the n numbers exceeds 45

In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46

thank you gmanjesh. it seems like a quicker approach. +1 for you

Re: CONSECUTIVE INTEGERS.. [#permalink]
14 Aug 2009, 17:05

gmanjesh wrote:

We can pick numbers as follows

Case 1) divide the number by 2 and play around with the numbers around it to get the sum Case 2) divide the number by 3 and play around with the numbers around it to get the sum Case 3) divide the number by 4 and play around with the numbers around it to get the sum Case 4) divide the number by 5 and play around with the numbers around it to get the sum and go on ... till you realize that sum of the n numbers exceeds 45

In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46

sum of consecutive integers=n/2(2a+n-1) where a is first term of series in the above equation we know all the terms except a and we need to find value of n S1 and S2 do not give value of "a" directly or indirectly. even after combining them we cant find value of "a". hence both statements are insufficient. Ans=E

since we dont know whether the first term of the sequence is 1 , we cant use the formula n(n+1)/2 formula bcoz that is sum of n consecutive natural numbers starting from 1

so using the formula n/2 (a+l), where l is the last term and a is the first term we get n/2(a+l) = 45 n(a+l)=90.....A1. 10*9 18*5 6*15 2*45 30*3 now statement 1 says n is even so all the above cases are possible but only 2*45 actually satisfies the condition. then A should be good enough.

2 CASE ) n < 9 only bottom three cases are possible But again here 2*45 is the only option which gives n=2 and terms will be 22 +23. no other soluton satisfies condition A1 ...hence B is also sufficient

I KNOW THAT I AM MAKING SOME MISTAKE BUT WHERE....PLS HELP

The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9

In how many ways can you write n consecutive integers such that their sum is 45? n = 1 -> 45 n = 2 -> 22, 23 n = 3 -> 14, 15, 16 n = 4 X n = 5 -> 7, 8, 9, 10, 11 n = 6 -> 5, 6, 7, 8, 9, 10 n = 7 X n = 8 X n = 9 -> 1, 2, 3, 4, 5, 6, 7, 8, 9

n can be 2 or 6 hence statement 1 is not sufficient. n can take many values less than 9 hence statement 2 is not sufficient. Since n can take values 2 or 6 which are even and less than 9, both statements together are not sufficient. Answer (E)

Now, the interesting thing is how to get these values of n. How do I know which values can n take and which can it not? Its pretty easy really. Follow my thought here.

"Of course n can be 1.

n can be 2 since when I divide 45 by 2, I get 22.5. So 2*22.5 is 45 so I have to find 2 consecutive integers whose mean is 22.5. The integers are obviously 22 and 23.

When I divide 45 by 3, I get 15. So I need 3 consecutive integers whose mean is 15. They are 14, 15, 16

When I divide 45 by 4, I get 11.25. Do I have 4 consecutive integers such that their mean is 11.25? No because mean of consecutive integers is always an integer or of the form n.5

When I divide 45 by 5, I get 9 so I need 5 consecutive integers whose mean is 9. They must be 7, 8, 9, 10, 11

and so on....."

Obviously, I just need to focus on getting 2 even values of n. So I check for 2, 4, 6 and I would know that answer is (E). _________________