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Three consecutive multiples of 3 have a sum of 36.

What is the greatest number?

 Nov 10, 2016

Best Answer 

 #2
avatar+121064 
+5

Let 3n be the first number, 3n + 3  be the second number and 3n + 6 be the third number....then....

 

3n + [3n+ 3] + [ 3n + 6]  = 36    simplify

 

9n +  9  = 36     subtract 9 from both sides 

 

9n =  27    divide both sides by 9

 

n = 3

 

So.....the first number 3(3)   = 9     tjhe second = 3(3) + 3  =  12     and the last = 3(3) + 6  = 15

 

And   9 + 12 + 15 =  36

 

 

So...the greatest number is 15

 

 

cool cool cool

 Nov 10, 2016
 #1
avatar+7188 
+5

Set up an equation.  Your equation should look like this: n+(n+3)+(n+6)=36

 

n is the smallest number,  therefore n+6 is the largest.

 

To simplify your equation,  it looks like:  3n+9=36

 

Subtract 9 from both sides.

 

3n+9-9=36-9

 

3n=27

 

Now divide both sides by 3.

 

3n/3=27/3

 

n=9

 

The smallest number is 9.

 

To find the largest number,  add 6 to n.

 

9 + 6 = 15

 

The greatest number is 15.

 

The middle number is 9 + 3 = 12.  The middle number is 12.

 

We can go back and check our work.

 

9 + 12 + 15 = 36

 

9,  12,  and 15 are consecutive multiples of 3.

 

Therefore,  your answer is 15.

 

@-Happy7

 Nov 10, 2016
 #2
avatar+121064 
+5
Best Answer

Let 3n be the first number, 3n + 3  be the second number and 3n + 6 be the third number....then....

 

3n + [3n+ 3] + [ 3n + 6]  = 36    simplify

 

9n +  9  = 36     subtract 9 from both sides 

 

9n =  27    divide both sides by 9

 

n = 3

 

So.....the first number 3(3)   = 9     tjhe second = 3(3) + 3  =  12     and the last = 3(3) + 6  = 15

 

And   9 + 12 + 15 =  36

 

 

So...the greatest number is 15

 

 

cool cool cool

CPhill Nov 10, 2016
 #3
avatar+121064 
+5

OK, H7....quit showing off...you're making me look bad....LOL!!!!

 

{ Actually....you're getting pretty good at this stuff ]

 

 

cool cool cool

 Nov 10, 2016
 #4
avatar+1313 
-4

Thanks a lot guys!

 Nov 10, 2016

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