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When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu

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When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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New post Updated on: 09 Feb 2018, 01:34
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When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively?

A. 3*119*120
B. 6*119*120
C. 3*118*119
D. 6*120*121
E. 3*120*121


*A solution will be posted in two days.

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Originally posted by MathRevolution on 11 Jan 2016, 16:54.
Last edited by MathRevolution on 09 Feb 2018, 01:34, edited 1 time in total.
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Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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New post 11 Jan 2016, 17:02
MathRevolution wrote:
When 1+2+..........+n = n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively?

A. 3*119*120
B. 6*119*120
C. 3*118*119
D. 6*120*121
E. 3*120*121


We want the sum of: 6 + 12 + 18 + 24 + ... + 708 + 714
Factor out a 6 to get: 6(1 + 2 + 3 + 4 + ... + 118 + 119)
Apply given formula to get: 6[(119)(119 + 1)/2]
Simplify: 6[(119)(120)/2]
Simplify more to get: (3)(119)(120)

Answer: A

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Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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New post 13 Jan 2016, 08:25
1
MathRevolution wrote:
When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively?

A. 3*119*120
B. 6*119*120
C. 3*118*119
D. 6*120*121
E. 3*120*121

*A solution will be posted in two days.


f(n) = 1 + 2 + ... + n = n(n+1)/2

714 = 6 x 119 and multiples of 6 to 714 include: 6x1, 6x2, 6x3, ....6x119

Sum of all multiples = 6 x f(119) = 6 x 119 x 120/2 = 3x119x120

Answer is A
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When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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New post Updated on: 09 Feb 2018, 14:13
When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively?

A. 3*119*120
B. 6*119*120
C. 3*118*119
D. 6*120*121
E. 3*120*121

-> 6+12+18+….+708+714=6(1+2+3+…+118+119)

=6[119(119+1)]/2=3*119*120.

Therefore, the answer is A.
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Originally posted by MathRevolution on 13 Jan 2016, 19:06.
Last edited by MathRevolution on 09 Feb 2018, 14:13, edited 2 times in total.
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Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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New post 08 Sep 2017, 10:17
First lets calculate no of terms there are from 6 to 714 that are multiples of 6
So
general formula for counting in steps of x is ( LT-FT)/x
If both ends are to be included add+1 or if both ends are excluded -1. If only one end taken then formula is as is.
Counting in steps of 6 We can write no terms as{ (714-6)/6}+1 = 119
Now since the seqn. will be in AP sum is 119 *{(6+714) /2}
= 119 *360= 3*119*120
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Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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New post 09 Sep 2017, 16:15
1
MathRevolution wrote:
When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively?

A. 3*119*120 B. 6*119*120 C. 3*118*119 D. 6*120*121 E. 3*120*121

*A solution will be posted in two days.

Half the challenge might be wading through answers as formatted.

For sums of multiples of sequences, I do not use factoring or n(n-1)/2.

I used the general formula for summing [multiples of] an arithmetic sequence, then found the value that matched my answer.

Formula for sum of multiples of arithmetic sequences:

(Average)(# of terms), where

Average = \(\frac{FirstTerm + Last Term}{2}\) and

# of Terms: \(\frac{Last - First}{Increment}\) + 1

Average: \(\frac{6 + 714}{2}\) = 360

# of Terms: \(\frac{708}{6}\) = 118 + 1 = 119

Sum of multiples of 6, from 6 to 714 inclusive, is (360)(119)

Check answers (and toss out B and D immediately, they're huge, and C, which does not end in a zero).

Answer A is a match.
3*119*120 = (119)(360)

ANSWER A
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Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu   [#permalink] 09 Sep 2017, 16:15
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