Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 21 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?
+118608 There are not any such integers.
But the easiest way to solve it is to let the integers be x, x+1, x+2, x+3, x+4, x+5, x+6
now do the equation bit. X equals a decimal, not an integer
Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 21 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?
The consecutive integers numbers are (x), (x+1), (x+2), (x+3), (x+4), (x+5), (x+6)
I came back and changed "consecutive integers" to "consecutive numbers"
because the solution reveals that the value of x is/was not an integer.
Their sum is 7x + 21
The problem says that this sum is 21 times the largest in the sequence.
So set it up like this 7x + 21 = (21)(x+6)
Work it out 7x + 21 = 21x + 126
Combine like terms 7x – 21x = 126 –21
–14x = 105
x = –7.5
7.5 is not an integer, so the postulate
fails, irrespective of polarity. No answer.
.
+135 The smallest integer that Pearl wrote down is 15.
Step-by-step explanation:
7 consecutive integers. They are:
a
a+1
a+2
a+3
a+4
a+5
a+6
Sum of the integers:
a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 = 7a + 21
6 times the largest of the seven integers.
The largest is a + 6. So
7a + 21 = 6(a + 6)
7a + 21 = 6a + 36
7a - 6a = 36 - 21
a = 15
The smallest integer that Pearl wrote down is 15.
