+0  
 
0
173
1
avatar

If the integer  is a perfect square, what is the sum of the digits of its square root?

 Jun 26, 2021

Best Answer 

 #1
avatar+2244 
+1

x^2 = 152AB1

sqrt(152001) = 389.873056263

So the number x is slightly greater than 389.873056263. 

Since 152001 ends with a 1, the number x has to end with a 1 or 9. 

391^2 = 152881

 

=^._.^=

 Jun 26, 2021
 #1
avatar+2244 
+1
Best Answer

x^2 = 152AB1

sqrt(152001) = 389.873056263

So the number x is slightly greater than 389.873056263. 

Since 152001 ends with a 1, the number x has to end with a 1 or 9. 

391^2 = 152881

 

=^._.^=

catmg Jun 26, 2021

15 Online Users

avatar