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Re: What is the sum of odd integers from 35 to 85, inclusive?
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03 Jul 2016, 11:08

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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
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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?

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Re: What is the sum of odd integers from 35 to 85, inclusive?
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01 Jul 2017, 11:18

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Notice that by adding pairs of numbers we get a sum of 120. We need to find the number of integers between 35 and 85 inclusive. The number of integers is \(\frac{(85-35)}{2}+1\) = 26

So we add 120 , 26 times to get 120*26, dividing this by 2 , we get 1560.
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What is the sum of odd integers from 35 to 85, inclusive?

Quote:

Number of odd integers = (85-35)/2 + 1 = 50/2 + 1 = 26

Can you explain why did you divide by 2? If I want to count no. of terms from 1 to 10, I would perform (Last term - First term) + 1. But is not above approach only valid for consecutive terms and sequence starting with 1?

Quote:

Sum of odd integers = (35+85)/2 * 26 = 60 * 26 = 1560

I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2. Let me know if we are on same page
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let me try to provide reply to your queries, Your query: Can you explain why did you divide by 2? Actually we divide by 2 because the difference between the consecutive odd numbers = 2.

In fact we have to count the multiple of any number say 7, multiple of 7 between 7 and 98 both inclusive. it will be (98-7)/7 +1 Always divide by common difference ( this is as per Arithmetic progression , AP) As per you "If I want to count no. of terms from 1 to 10, I would perform (Last term - First term) + 1." you are right here, the common difference = 1. so dividing by 1 is not necessary. Your Query :I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2. yes you are right, in fact, it is true for any set of equally spaced numbers (AP)

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

Quote:

Number of odd integers = (85-35)/2 + 1 = 50/2 + 1 = 26

Can you explain why did you divide by 2? If I want to count no. of terms from 1 to 10, I would perform (Last term - First term) + 1. But is not above approach only valid for consecutive terms and sequence starting with 1?

Quote:

Sum of odd integers = (35+85)/2 * 26 = 60 * 26 = 1560

I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2. Let me know if we are on same page

Add the numbers in pairs, starting from the outside and working towards the middle, we get: 35 + 37 + 39 + 41 + . . . . 79 + 81 + 83 + 85 = (35 + 85) + (37 + 83) + (39 + 81) + . . . = (120) + (120) + (120) + (120) + ..... = 120 x some integer

So, the desired sum must be DIVISIBLE by 120 Only one answer choice is divisible by 120

Answer: A

This observation is interesting but I would take a separate approach after the first two steps:-

Add the numbers in pairs, starting from the outside and working towards the middle, we get: 35 + 37 + 39 + 41 + . . . . 79 + 81 + 83 + 85 = (35 + 85) + (37 + 83) + (39 + 81) + . . . = (120) + (120) + (120) + (120) + ..... = 120 x some integer

Some integer is 13 because the average of extreme values is 60. So for every odd number from 35 to 60 there will be a pair for it from 60 to 85. Its easy to count that the number of pairs will be 13. therefore the answer is 120*13. [Only reason I mentioned this approach was because sometimes dividing 5 numbers could be time consuming. Counting the number of pairs to 13 was simple and straight forward)

Edit: Formating

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let me try to provide reply to your queries, Your query: Can you explain why did you divide by 2? Actually we divide by 2 because the difference between the consecutive odd numbers = 2.

In fact we have to count the multiple of any number say 7, multiple of 7 between 7 and 98 both inclusive. it will be (98-7)/7 +1 Always divide by common difference ( this is as per Arithmetic progression , AP) As per you "If I want to count no. of terms from 1 to 10, I would perform (Last term - First term) + 1." you are right here, the common difference = 1. so dividing by 1 is not necessary. Your Query :I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2. yes you are right, in fact, it is true for any set of equally spaced numbers (AP)

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

Quote:

Number of odd integers = (85-35)/2 + 1 = 50/2 + 1 = 26

Can you explain why did you divide by 2? If I want to count no. of terms from 1 to 10, I would perform (Last term - First term) + 1. But is not above approach only valid for consecutive terms and sequence starting with 1?

Quote:

Sum of odd integers = (35+85)/2 * 26 = 60 * 26 = 1560

I guess you used that for consecutive numbers, mean = median = (First term + last term) / 2. Let me know if we are on same page

I think you are missing one point here Always divide by common difference ( this is as per Arithmetic progression , AP) this is not true we have to divide by common difference of first and last multiple of the integer whose multiples are being found out in this case 7 and 98 are by chance multiples of 7 lets say multiples b/w 7 and 99 even then we do 98-7
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