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What is the sum of odd integers from 35 to 85, inclusive?

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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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New post 01 Aug 2019, 17:04
diegocml wrote:
Skywalker18 wrote:
Bunuel wrote:
What is the sum of odd integers from 35 to 85, inclusive?

A) 1,560
B) 1,500
C) 1,240
D) 1,120
E) 1,100


Number of odd integers = (85-35)/2 + 1
= 50/2 + 1
= 26
Sum of odd integers = (35+85)/2 * 26
= 60 * 26
= 1560
Answer A


Hi Skywalker18,

I do understand the following on this exercise:

1. We're dealing with consecutive integers so I can find the average by: \(\cfrac { First\quad Number+Last\quad Number }{ 2 }\)
2. To find the range in a set of numbers for which both end points are inclusive, we: \(\left( Hightest\quad Number-Lowest\quad Number \right) +1\)

I don't understand why you can simply divide the range by 2 in \(\cfrac { \left( Hightest\quad Number-Lowest\quad Number \right) }{ 2 } +1\) to find the number of odd numbers. Would it be the same if I wanted to find the number of even numbers?

Thanks'


In case anyone else finds the highlighted part confusing:

If we wanted to take the even numbers we would have to change the set to NOT include 35 and 85 (since they are odd)
The calculation would now be [(Last = 84) - (First = 36) ]/ (space between each number = 2) + 1 (because inclusive is mentioned)
So, 84-36/2 + 1 = 48/2 + 1 = 24 + 1 = 25, not 26.
Inclusive means we add +1 to include a number we would otherwise cut off in the calculation, regardless of E or O.
e.g. 1,2,3,4,5, how many numbers? 5 by counting, if we use formula (5-1)/1 = 4, we are missing a number if we don't add +1.

Also, the other point is about symmetry. If we start with an E and end in an O (or vice versa) there will be the same amount of E and O, e.g. 0, 1, 2, 3, 4, 5 (3 E, 3 O)
BUT, if we start with an E and end in an E (or vice versa), then there is 1 more E, e.g. 0, 1, 2, 3, 4, 5, 6 (4 E, 3 O)
This info can be useful for some combinations/probability problems.
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Re: What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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New post 09 Aug 2019, 23:11
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My approach: from 35 to 85 inclusive, there is (( 85-35)+1)=51 integers. As the number of odd integers is 1 more than even integers (because 35 and 85 both are odd number, so the number of odd would be 1 more than even number), the total number of odd integers would be 26 (even integer 25). We know the formulae of sum of evenly spaced numbers is : (((85+35)/2)* number of terms 26)=1560, which is the sum of odd integers from 35 to 85.
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What is the sum of odd integers from 35 to 85, inclusive?  [#permalink]

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New post 23 Oct 2019, 13:50
Sum of first n consecutive positive odd integers: n^2
from 1-35=17 odd integers
from 1-85= 43 odd integers (85 inclusive)
=43^2 - 17^2
=1560
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What is the sum of odd integers from 35 to 85, inclusive?   [#permalink] 23 Oct 2019, 13:50

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