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# Equations

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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?

May 2, 2023

#1
+118616
+1

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

May 2, 2023
#2
+1

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.

.

May 2, 2023
#3
+135
+1

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.

May 2, 2023
#4
+1

You said "6 times the largest of the seven integers"

The problem said "21 times the largest of the seven integers"

You solved a different problem from the problem that was posed.

.

Guest May 2, 2023