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

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by Max@Math Revolution » Wed Jan 13, 2016 6:32 pm
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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by Brent@GMATPrepNow » Wed Jan 13, 2016 6:37 pm
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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Brent
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by chetan.sharma » Thu Jan 14, 2016 4:12 am
Max@Math Revolution 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.
Hi,
two points..
1) the formula for sum of first n positive integer is required to be known by the students and I do not think they would give this in the Q stem..
2) the solution can be found even without that by taking average and multiplying by total numbers..

since the method by formula is already touched, I'll do it by the other average method..

average of all multiples of 6 till 714 is (6+714)/2..
this is so because 714 is also multiple of 6, otherwise we would have taken the last multiple of 6 before 714..
total number of multiples=714/6=119..
so our answer is 119*720/2 is the same as 119*3*120..
ans A
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by Max@Math Revolution » Mon Jan 18, 2016 8:58 pm
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+4+...+118+119)=6[119(119+1)]/2=3*119*120.
Therefore, the answer is A.
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