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The numbers $24^2 = 576$ and $56^2 = 3136$ are examples of perfect squares that have a units digits of $6.$

 

If the units digit of a perfect square is $5,$ then what are the possible values of the tens digit?

 Jul 1, 2024
 #1
avatar+867 
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The possible values of the tens digit are 0, 1, 2, 3, 5, 6, 7, and 9.

 Jul 2, 2024
 #2
avatar+129739 
+1

If 5 is the units digit of a perfect square.....the original squared integer must end in 5

 

Let the other part of the original integer be some non-negative  multiple of 10...call this part "a"

 

Then

 

(a + 5) (a + 5)  = 

 

a^2 + 10a + 25

 

a^2 will either = 0  or it will end in at least two 0's

 

10a  will = 0  or it will end in at least two 0's

 

Then the tens digit of the perfect square wil always = 2

 

cool cool cool

 Jul 2, 2024
edited by CPhill  Jul 2, 2024

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