Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 28/5 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?

Answer 1

Let the smallest integer that Pearl wrote down be x. The next six consecutive integers would be x+1, x+2, x+3, x+4, x+5, and x+6.

The sum of these seven integers is x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) = 7x + 21.

According to the problem, this sum is equal to 28/5 times the largest integer, which is x+6. Therefore, we have:

7x + 21 = (28/5)(x+6)

Multiplying both sides by 5 to get rid of the fraction, we get:

35x + 105 = 28(x+6)
35x + 105 = 28x + 168
7x = 63
x = 9

Therefore, the smallest integer that Pearl wrote down is 9.