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Problem Solving Pack 4, Question 2) For every positive integer n...
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19 Nov 2015, 16:37
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QUANT 4PACK SERIES Problem Solving Pack 4 Question 2 For every positive integer n...For every positive integer n that is greater than 1, the function g(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The g(n) could have any of the following units digits except…? A) 1 B) 2 C) 4 D) 6 E) 9 48 Hour Window Answer & Explanation WindowEarn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4Pack seriesScroll Down For Official Explanation
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Re: Problem Solving Pack 4, Question 2) For every positive integer n...
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19 Nov 2015, 18:08
For every positive integer n that is greater than 1, the function g(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The g(n) could have any of the following units digits except…?
Used brute force on this one to calculate. Not sure if there is a quicker way.
1+3=4 C is out 1+3+....15= 81 A is out 1+3+5+7=16 D is out 1+3+5=9 E i out
Answer: B



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Re: Problem Solving Pack 4, Question 2) For every positive integer n...
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19 Nov 2015, 20:36
EMPOWERgmatRichC wrote: For every positive integer n that is greater than 1, the function g(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The g(n) could have any of the following units digits except…?
A) 1 B) 2 C) 4 D) 6 E) 9
Notice the pattern: g(2) = 1g(3) = 1 + 3 = 4g(4) = 1 + 3 = 4g(5) = 1 + 3 + 5 = 9g(6) = 1 + 3 + 5 = 9g(7) = 1 + 3 + 5 + 7 = 16At this point, we've already eliminated A, C, D and E. So the answer is B. Alternatively, we might notice that g(n) is always the square of an integer. Since there is no integer that, when squared, has a units digit of 2, the answer must be B Cheers, Brent
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Re: Problem Solving Pack 4, Question 2) For every positive integer n...
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19 Nov 2015, 20:44
Sum of all odd numbers from 1 to n = n^2
Out of all those answers, only 2 is not the digit of any perfect square. So the answer is B.



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Re: Problem Solving Pack 4, Question 2) For every positive integer n...
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23 Nov 2015, 23:54
EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Problem Solving Pack 4 Question 2 For every positive integer n...
For every positive integer n that is greater than 1, the function g(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The g(n) could have any of the following units digits except…?
A) 1 B) 2 C) 4 D) 6 E) 9
Hi All, GMAT Quant questions are built around patterns  whether you immediately spot (or know) the pattern or not. As such, even if you don't know the exact patterns involved, you can often PROVE what the pattern is by TESTing VALUES and doing some basic arithmetic. Here, we're told that n is a positive integer greater than 1 and that the function g(n) is defined to be the sum of all of the ODD integers from 1 to n, inclusive. We're asked to find the digit that CAN'T be the unit's digit of any potential calculation that could be done using this function. Rather than stare at the screen or do math in our heads, let's TEST VALUES and seek out the pattern. Since n is GREATER than 1, let's start with 2... g(2) = 1 So the unit's digit COULD be 1. Eliminate Answer A. g(3) = 1 + 3 = 4 So the unit's digit COULD be 4. Eliminate Answer C. g(5) = 1 + 3 + 5 = 9 So the unit's digit COULD be 9. Eliminate Answer E. g(7) = 1 + 3 + 5 + 7 = 16 So the unit's digit COULD be 6. Eliminate Answer D. With 4 answers eliminated, there's just one answer left... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Problem Solving Pack 4, Question 2) For every positive integer n...
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01 Jul 2016, 22:50
EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Problem Solving Pack 4 Question 2 For every positive integer n...For every positive integer n that is greater than 1, the function g(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The g(n) could have any of the following units digits except…? A) 1 B) 2 C) 4 D) 6 E) 9 48 Hour Window Answer & Explanation WindowEarn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4Pack seriesScroll Down For Official Explanation Sum of the first n positive integers upto n is given by \(n^2\) now these can be either even number of odd numbers or odd number of odd numbers. Thus the question is "Which of the following cannot be the units digit of a perfect square". Answer (B).
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Re: Problem Solving Pack 4, Question 2) For every positive integer n...
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21 Oct 2016, 06:18
EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Problem Solving Pack 4 Question 2 For every positive integer n...
For every positive integer n that is greater than 1, the function g(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The g(n) could have any of the following units digits except…?
A) 1 B) 2 C) 4 D) 6 E) 9
1+3 = 4 > C is out 4+5 = 9 > E is out 9+7 = 16 > D is out 16+9 = 25. 25+11 = 36 36+13 = 49 49+15 = 64 64+17 = 81  A is out only left  B!



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Re: Problem Solving Pack 4, Question 2) For every positive integer n...
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21 Oct 2016, 08:36
EMPOWERgmatRichC wrote: For every positive integer n that is greater than 1, the function g(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The g(n) could have any of the following units digits except…?
A) 1 B) 2 C) 4 D) 6 E) 9
\( No need of even a single calculation\) Function \(g(n) =\) \(n^2\) { Sum of odd integers } Since \(n^2\) is a square number it must end in 1 , 2 , 4 , 6 , 9 and 0 Since Units digit of \(n^2\) can not be 2, our answer will be (B)
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Re: Problem Solving Pack 4, Question 2) For every positive integer n...
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29 Jul 2018, 20:58
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