The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is odd
(2) n >= 9
(1) n is odd
(2) n >= 9
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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.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.
My approach is sum of n numbers(starting from 1) is n(n+1)/2.shashank.ism wrote: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.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.
If there would have been nos. like 469 or 6779 instead of 45 what wll you do??
1)for n=3 say a-1 ,a,a+1gmatnmein2010 wrote:The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is odd
(2) n >= 9
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