Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 8 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?
n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n + 6 ==8 * [n + 6], solve for n
7n + 16 =8n + 48
7n - 8n ==48 - 16
- n = 32 or:
n == - 32 - the smallest n
Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 8 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?
Since the integers are consecutive, each integer's value will be one more than the integer before.
let x be the first integer
then the seven integers are (x) + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6)
and their sum is 7x + 21
we want this to be 8 times the largest number
so 7x + 21 = 8 times (x+6)
7x + 21 = 8x + 48
7x – 8x = 48 – 21
– x = 27
x = –27
check answer by plugging
into the original equality 7x + 21 = 8 times (x+6)
(7)(–27) + 21 = (8)(–27) + 48
–189 + 21 = –216 + 48
–168 = –168 looks good
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