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# Problem Solving Pack 4, Question 2) For every positive integer n...

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Re: Problem Solving Pack 4, Question 2) For every positive integer n... [#permalink]
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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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Re: Problem Solving Pack 4, Question 2) For every positive integer n... [#permalink]
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EMPOWERgmatRichC wrote:
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

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.

g(2) = 1
So the unit's digit COULD be 1.

g(3) = 1 + 3 = 4
So the unit's digit COULD be 4.

g(5) = 1 + 3 + 5 = 9
So the unit's digit COULD be 9.

g(7) = 1 + 3 + 5 + 7 = 16
So the unit's digit COULD be 6.

GMAT assassins aren't born, they're made,
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Re: Problem Solving Pack 4, Question 2) For every positive integer n... [#permalink]
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EMPOWERgmatRichC wrote:
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
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

Scroll 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".

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Re: Problem Solving Pack 4, Question 2) For every positive integer n... [#permalink]
EMPOWERgmatRichC wrote:
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

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... [#permalink]
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... [#permalink]
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Re: Problem Solving Pack 4, Question 2) For every positive integer n... [#permalink]
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