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The sum of the positive integers from 1 through n can be cal
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14 Apr 2018, 02:31
14 Apr 2018, 02:31
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84% (01:49) correct
15% (01:07) wrong based on 85 sessions
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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
Re: The sum of the positive integers from 1 through n can be cal
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14 Apr 2018, 22:48
14 Apr 2018, 22:48
3
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
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
Re: The sum of the positive integers from 1 through n can be cal
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15 Apr 2018, 12:20
15 Apr 2018, 12:20
4
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
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
