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# The sum of all the multiples of 6 between –126 and 342, incl

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The sum of all the multiples of 6 between –126 and 342, incl [#permalink]  15 Sep 2017, 08:11
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Question Stats:

90% (01:01) correct 9% (02:01) wrong based on 11 sessions

 Quantity A Quantity B The sum of all the multiples of 6 between –126and 342, inclusive 8,502

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

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Re: The sum of all the multiples of 6 between –126 and 342, incl [#permalink]  06 Dec 2017, 11:09
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If we look closely, we only have to sum Integers from 127 to 342, incl. as the negative 126 Integers will cancel out the 126 positive Integers.

=> We have an evenly spaced sequence (Multiples of 6). Hence we can take the first value of the sequence + the last value of the sequence and divide the sum by 2 to obtain the mean of the sequence. We multiply the mean times the number of Integers.

(132+342)/(2) * ((342-132/6) +1)) = 357 * 36 = 8532 Hence A is the answer
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Re: The sum of all the multiples of 6 between –126 and 342, incl [#permalink]  12 Dec 2017, 22:05
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Carcass wrote:

 Quantity A Quantity B The sum of all the multiples of 6 between –126and 342, inclusive 8,502

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Let me explain in a broader way-

Now we need the sum of multiples of 6 from -126 to 342.

SO to add these multiples we can find that number from -126 to +126 will cancel out, so we only need to find the sum of multiples of 6 from number 127 to 342

NOw there is a formula to find the number of multiples;

Multiple of x in the range = $$\frac{(last multiple of "x" in the range - first multiple of "x" in the range)}{x}$$ +1

now we need to find the first multple of 6 in the range from 127 to 342.

The first multiple is 6 *22 = 132

and the last multiple is 6 * 57 =342

Now multple of 6 in the range from 127 to 342 =$$\frac{(342 - 132)}{6} + 1$$

= 36 numbers which are the multiple of 6 in the range from 127 to 342

Now

here we can also write as 6*22 + 6*23 + 6*24 + ...+6*54

or 6 (22 + 23 + 24 + ...)

SInce there are total of 36 numbers so the mean of the number * total number will give us the sum i.e

Mean of the numbers will be the 17th and 18th number i.e = 39 and 40 (in the range from 22 to 57)
So the sum (22 + 23 + 24 + ....+ 57)= $$\frac{(39 + 40)}{2} * 36$$ = 1422

Therefore the total sum of multiple of 6 from the range 127 to 342 = 1422 * 6 =8532

* Note
There is a formula to find the sum of 'X' consecutive number

if it is odd then then sum will be the = middle number * X

if it is even then the sum will be = mean of the numbers in the range * X
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Re: The sum of all the multiples of 6 between –126 and 342, incl   [#permalink] 12 Dec 2017, 22:05
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