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+0  
 
0
86
5
avatar+526 

Can anyone solve this?

 Mar 14, 2019
 #1
avatar
0

Sorry, didn't see that!.

 Mar 14, 2019
edited by Guest  Mar 14, 2019
 #2
avatar+101870 
0

0   isn't positive

 

cool cool cool

CPhill  Mar 14, 2019
 #3
avatar+101870 
+3

Here's my best attempt

 

Assume that the only two integers are 1 and 3.......and.....the number of 3s is one more than the number of 1s 

 

So....

 

[3(n + 1) + 1(n) ]

______________     <   2.1

        2n + 1

 

4n + 3  <     4.2n + 2.1

 

.9  < .2n

 

9/2 < n

 

n = 5 

 

 

 1 1 1 1 1 3 3 3 3 3 3

 

We need 11 integers

 

 

cool  cool cool

 Mar 14, 2019
 #4
avatar+5226 
+3

I second CPhil's answer

Rom  Mar 14, 2019
 #5
avatar+526 
0

Thank you for your help!!! 11 is the correct answer.

dgfgrafgdfge111  Mar 14, 2019

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