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by papgust » Mon Feb 08, 2010 5:45 am
I would go with B here

1. Truly insufficient.

Take nos 1 to 9. sum equals 45.
Take nos 7 to 11. sum equals 45

2. n >= 9

Nos only with 1 to 9 sum upto 45. Greater than 9 integers will not get you the sum 45.

Sufficient.
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by shashank.ism » Mon Feb 08, 2010 6:32 am
papgust wrote:I would go with B here

1. Truly insufficient.

Take nos 1 to 9. sum equals 45.
Take nos 7 to 11. sum equals 45

2. n >= 9

Nos only with 1 to 9 sum upto 45. Greater than 9 integers will not get you the sum 45.

Sufficient.
What would be our approach in this problem. How u calculated that sum of 1 to 9 =45 and 7 to 11 = 45.. There must be some other way to solve this problem.

If there would have been nos. like 469 or 6779 instead of 45 what wll you do??
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by harsh.champ » Mon Feb 08, 2010 6:52 am
shashank.ism wrote:
papgust wrote:I would go with B here

1. Truly insufficient.

Take nos 1 to 9. sum equals 45.
Take nos 7 to 11. sum equals 45

2. n >= 9

Nos only with 1 to 9 sum upto 45. Greater than 9 integers will not get you the sum 45.

Sufficient.
What would be our approach in this problem. How u calculated that sum of 1 to 9 =45 and 7 to 11 = 45.. There must be some other way to solve this problem.

If there would have been nos. like 469 or 6779 instead of 45 what wll you do??
My approach is sum of n numbers(starting from 1) is n(n+1)/2.

statement 2:-only possible if n=9 [sum is 45 as shown by me above]

Hey shashank,
Hope u have seen my upper formula .
So,sum from 7 to 11 can be done using:-Sum of 1st 11 no.s - -Sum of 1st 6 no.s
=11(11 + 1)/2 - 6(6 + 1)/2 =45.

This is the proper,formal and a very short method.[It doesn't take too much time if u know the formula]
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by thephoenix » Mon Feb 08, 2010 10:43 am
gmatnmein2010 wrote:The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is odd
(2) n >= 9
1)for n=3 say a-1 ,a,a+1
a-1+a+a+1=3a=45--->a=15 hence seq is 14,15,17

for n=5
a-2,a-1,a,a+1,a+2
sum=5a=45--->a=9
hence seq is 7,8,9,10,11

we can have 1,2,3,4,5,6,7,8,9 n=9 also
hence n varies and therefore insuff
2)consecutive positive integers max for n is 9: 1+2+3+...+9=45.
when n>9=say 10
then first term becomes zero.
but we are given that all terms are positive)
So only case n=9 is possible.
hence Sufficient.

hence B
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