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EMPOWERgmatRichC
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For every positive integer n that is greater than 1, the function g(n) 19 Nov 2015, 17:37
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B
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67% (01:58) correct 33% (01:56) wrong based on 429 sessions

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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

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QUANT 4-PACK SERIES Problem Solving Pack 4 Question 2 For every positive integer n...


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B
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peachfuzz
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For every positive integer n that is greater than 1, the function g(n) 19 Nov 2015, 19:08
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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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For every positive integer n that is greater than 1, the function g(n) 19 Nov 2015, 21:36
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EMPOWERgmatRichC

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
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Notice the pattern:
g(2) = 1
g(3) = 1 + 3 = 4
g(4) = 1 + 3 = 4
g(5) = 1 + 3 + 5 = 9
g(6) = 1 + 3 + 5 = 9
g(7) = 1 + 3 + 5 + 7 = 16

At 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

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For every positive integer n that is greater than 1, the function g(n) 19 Nov 2015, 21:44
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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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For every positive integer n that is greater than 1, the function g(n) 24 Nov 2015, 00:54
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QUANT 4-PACK 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
Show more

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:
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B

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For every positive integer n that is greater than 1, the function g(n) 01 Jul 2016, 23:50
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EMPOWERgmatRichC
QUANT 4-PACK 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


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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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For every positive integer n that is greater than 1, the function g(n) 21 Oct 2016, 07:18
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QUANT 4-PACK 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
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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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For every positive integer n that is greater than 1, the function g(n) 21 Oct 2016, 09:36
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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

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\(\\
No need of even a single calculation\)

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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For every positive integer n that is greater than 1, the function g(n) 26 Sep 2024, 01:35
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the sum of the odd integers from 1 to n --> g(n) --> is equal to n^2
we have to check if there are perfect squares that end in one of the digits provided.
thus, by checking the values of n^2, we can conclude that 2 is the only digit that cannot be unit digit


EMPOWERgmatRichC
QUANT 4-PACK 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 Window
Earn 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 4-Pack series

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