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Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 4/3 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?

 Jun 22, 2024

Best Answer 

 #1
avatar+10 
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Let the numbers be \(x-3,x-2,x-1,x,x+1,x+2,x+3\)
Their sum is
 \(7x=\frac{4}{3}(x+3)\)


Solving we get:
\(x=\frac{12}{17}\) 
The largest number is \(x+3=\frac{12}{17}+\frac{3}{1}=\boxed{\frac{63}{17}}\)
Which is not an integer and the question is flawed. 

 Jun 22, 2024
 #1
avatar+10 
+1
Best Answer

Let the numbers be \(x-3,x-2,x-1,x,x+1,x+2,x+3\)
Their sum is
 \(7x=\frac{4}{3}(x+3)\)


Solving we get:
\(x=\frac{12}{17}\) 
The largest number is \(x+3=\frac{12}{17}+\frac{3}{1}=\boxed{\frac{63}{17}}\)
Which is not an integer and the question is flawed. 

EnormousBighead Jun 22, 2024

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