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by Brent@GMATPrepNow » Thu Sep 12, 2013 5:44 am
twinkzz wrote:the sum of four consecutive multiples of 16 is 2048 more than the second number. what is the third number????
Four consecutive multiples of 16, can be written as 16x, 16(x+1), 16(x+2), 16(x+3), where x is some integer.

The sum of four consecutive multiples of 16 . . .
Sum = 16x + 16(x+1) + 16(x+2) + 16(x+3)
= 64x + 96

. . . is 2048 more than the second number
So, 64x + 96 is 2048 greater than 16(x+1)
We can write: 64x + 96 - 16(x+1) = 2048
Expand: 64x + 96 - (16x + 16) = 2048
Simplify: 48x + 80 = 2048
Solve: x = 41

So, the 1st multiple of 16 = (41)(16) = 656
The 2nd multiple of 16 = 656 + 16 = 672
The 3rd multiple of 16 = 672 + 16 = 688

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by GMATGuruNY » Thu Sep 12, 2013 7:20 am
twinkzz wrote:the sum of four consecutive multiples of 16 is 2048 more than the second number. what is the third number????


plzz answer :?
Alternate approach:

Let the third number = x.
Then the second number = x-16.

When integers are EVENLY SPACED, sum = (number of integers)(median).
Here, the median is HALFWAY between the third number and the second number: x-8.
Thus:
Sum = (number of integers)(median) = 4(x-8) = 4x-32.

Since the difference between the sum and the second number is 2048, we get:
(4x-32) - (x-16) = 2048
3x-16 = 2048
3x = 2064
x = 688.
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