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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # The sum of the positive integers from 1 through n can be cal  Question banks Downloads My Bookmarks Reviews Important topics
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Founder  Joined: 18 Apr 2015
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The sum of the positive integers from 1 through n can be cal [#permalink]
Expert's post 00:00

Question Stats: 82% (01:48) correct 17% (00:06) wrong based on 23 sessions

The sum of the positive integers from 1 through n can be calculated by the formula $$\frac{n(n + 1)}{2}$$

 Quantity A Quantity B The sum of the multiples of 6 between 0 and 100 The sum of the multiples of 8 between 0 and 100

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
[Reveal] Spoiler: OA

_________________ Director  Joined: 07 Jan 2018
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Re: The sum of the positive integers from 1 through n can be cal [#permalink]
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Find the value of $$n$$ for each quantity

For A: the numbers are multiples of $$6$$; 1st multiple = $$0$$ and the last multiple below $$100 = 96$$
value of $$n = \frac{(96-0)}{6} + 1 = 17$$ (since both the numbers are divisible by $$6$$)

sum $$= 17 * \frac{96}{2}= 816$$

For B: the numbers are multiples of$$8$$; 1st multiple $$= 0$$ and the last multiple below $$100 = 96$$
value of $$n = \frac{96-0}{8} + 1 = 13$$ (since both the numbers are divisible by $$6$$)

sum = $$13 * 96/2 = 624$$

clearly $$A$$ is bigger
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes Intern Joined: 14 Apr 2018
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Re: The sum of the positive integers from 1 through n can be cal [#permalink]
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amorphous wrote:
Find the value of $$n$$ for each quantity

For A: the numbers are multiples of $$6$$; 1st multiple = $$0$$ and the last multiple below $$100 = 96$$
value of $$n = \frac{(96-0)}{6} + 1 = 17$$ (since both the numbers are divisible by $$6$$)

From formula, sum $$= 17 * \frac{18}{2} = 17*9$$

For B: the numbers are multiples of$$8$$; 1st multiple $$= 0$$ and the last multiple below $$100 = 96$$
value of $$n = \frac{96-0}{8} + 1 = 13$$ (since both the numbers are divisible by $$6$$)

From formula, sum = $$13 * \frac{14}{2} = 13*7$$

clearly $$A$$ is bigger

Hi, I found that the you made a mistake with the sum above.
I have a different solution for this question:
- The first multiple of 6 is: 0 = 6*0
- The second multiple of 6 is: 6 =6*1
- The third multiple of 6 is: 12=6*2
............
- The seventeenth multiple of 6 is: 96=6*16
--> the sum of all multiples of 6 is: 6*(0+1+....+16) = 6*(1+...+16)=6*(16*17)/2=8*6*17
Similarly,
--> the sum of all multiples of 8 is: 8*(1+...+12)=8*(12*13)/2=8*6*13

So A is bigger
Director  Joined: 07 Jan 2018
Posts: 642
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Kudos [?]: 593 , given: 88

Re: The sum of the positive integers from 1 through n can be cal [#permalink]
halam wrote:
amorphous wrote:
Find the value of $$n$$ for each quantity

For A: the numbers are multiples of $$6$$; 1st multiple = $$0$$ and the last multiple below $$100 = 96$$
value of $$n = \frac{(96-0)}{6} + 1 = 17$$ (since both the numbers are divisible by $$6$$)

From formula, sum $$= 17 * \frac{18}{2} = 17*9$$

For B: the numbers are multiples of$$8$$; 1st multiple $$= 0$$ and the last multiple below $$100 = 96$$
value of $$n = \frac{96-0}{8} + 1 = 13$$ (since both the numbers are divisible by $$6$$)

From formula, sum = $$13 * \frac{14}{2} = 13*7$$

clearly $$A$$ is bigger

Hi, I found that the you made a mistake with the sum above.
I have a different solution for this question:
- The first multiple of 6 is: 0 = 6*0
- The second multiple of 6 is: 6 =6*1
- The third multiple of 6 is: 12=6*2
............
- The seventeenth multiple of 6 is: 96=6*16
--> the sum of all multiples of 6 is: 6*(0+1+....+16) = 6*(1+...+16)=6*(16*17)/2=8*6*17
Similarly,
--> the sum of all multiples of 8 is: 8*(1+...+12)=8*(12*13)/2=8*6*13

So A is bigger

can you clarify what is the mistake?
Intern Joined: 14 Apr 2018
Posts: 7
Followers: 0

Kudos [?]: 3 , given: 5

Re: The sum of the positive integers from 1 through n can be cal [#permalink]
amorphous wrote:
halam wrote:
amorphous wrote:
Find the value of $$n$$ for each quantity

For A: the numbers are multiples of $$6$$; 1st multiple = $$0$$ and the last multiple below $$100 = 96$$
value of $$n = \frac{(96-0)}{6} + 1 = 17$$ (since both the numbers are divisible by $$6$$)

From formula, sum $$= 17 * \frac{18}{2} = 17*9$$

For B: the numbers are multiples of$$8$$; 1st multiple $$= 0$$ and the last multiple below $$100 = 96$$
value of $$n = \frac{96-0}{8} + 1 = 13$$ (since both the numbers are divisible by $$6$$)

From formula, sum = $$13 * \frac{14}{2} = 13*7$$

clearly $$A$$ is bigger

It is the sum, the number of multiples of 6 is not continuous from

Hi, I found that the you made a mistake with the sum above.
I have a different solution for this question:
- The first multiple of 6 is: 0 = 6*0
- The second multiple of 6 is: 6 =6*1
- The third multiple of 6 is: 12=6*2
............
- The seventeenth multiple of 6 is: 96=6*16
--> the sum of all multiples of 6 is: 6*(0+1+....+16) = 6*(1+...+16)=6*(16*17)/2=8*6*17
Similarly,
--> the sum of all multiples of 8 is: 8*(1+...+12)=8*(12*13)/2=8*6*13

So A is bigger

can you clarify what is the mistake?

The sum you come up with: sum of all multiples of 6 = (17 * 18)/2 = 17*9
Since multiples of 6 are not continuous integers from 1 to 17, you cannot calculate the sum as above.
Please let me know your thought. Director  Joined: 07 Jan 2018
Posts: 642
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Kudos [?]: 593  , given: 88

Re: The sum of the positive integers from 1 through n can be cal [#permalink]
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Yes you are correct the given formula is for continuous numbers.
We can use avg * number of terms or similar methods to reach the answer. I have modified the above solution
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes Re: The sum of the positive integers from 1 through n can be cal   [#permalink] 16 Apr 2018, 21:18
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