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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?

 Jun 13, 2024

Best Answer 

 #1
avatar+1252 
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We can set up an equation to help us solve this problem. Let's define a variable. 

Let n be the smallest  number. We have \(n + n+1 + n+2 ....... n+6 = 7n + 21\)

 

Now, this sequence adds up to 28/5 times the largest number, n + 6, so we can write the equation

\(7n+21=28/5(n+6)\)

 

From here, we just solve for n. We have

\(7n+21=\frac{28}{5}n + 168/5\\ 7/5n=63/5\)

 

Now, multiplying both sides by the reciprocol of 7/5, we get

\(n=\frac{63}{5} \cdot \frac{5}{7}=9\)

 

So 9 is our final answer!

 

Thanks! :)

 Jun 13, 2024
 #1
avatar+1252 
+1
Best Answer

We can set up an equation to help us solve this problem. Let's define a variable. 

Let n be the smallest  number. We have \(n + n+1 + n+2 ....... n+6 = 7n + 21\)

 

Now, this sequence adds up to 28/5 times the largest number, n + 6, so we can write the equation

\(7n+21=28/5(n+6)\)

 

From here, we just solve for n. We have

\(7n+21=\frac{28}{5}n + 168/5\\ 7/5n=63/5\)

 

Now, multiplying both sides by the reciprocol of 7/5, we get

\(n=\frac{63}{5} \cdot \frac{5}{7}=9\)

 

So 9 is our final answer!

 

Thanks! :)

NotThatSmart Jun 13, 2024

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