the sum of four consecutive multiples of 16 is 2048 more than the second number. what is the third number????
plzz answer
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Four consecutive multiples of 16, can be written as 16x, 16(x+1), 16(x+2), 16(x+3), where x is some integer.twinkzz wrote:the sum of four consecutive multiples of 16 is 2048 more than the second number. what is the third number????
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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Alternate approach:twinkzz wrote:the sum of four consecutive multiples of 16 is 2048 more than the second number. what is the third number????
plzz answer
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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