+0  
 
0
169
3
avatar+61 

Can someone help with this?

 May 20, 2021
 #2
avatar
+2

 

Any integer that ends with a 6 raised to any power results in an integer that ends with a 6. 

So 196213 ends with a 6.  

 

There's that, but the 3 is a little harder.  

 

Notice the last digit of the results of 3 raised to various powers.  

 

When "n" is:                0   1   2   3    4   5   6   7    8   9   10   11    12   13   14   15  

The last digit of 3n is:  1   3   9   7    1   3   9   7    1   3     9    7      1     3     9     7  

 

That's enough to show that every fourth group keeps repeating 1 3 9 7   

 

So divide 196 by 4 and you get 49 complete groups with no remainder. 

That means that 213196 ends with a 7.  

 

6 times 7 is 42 so ............ the last digit of 196213 times 213196 is a 2.   

 

If this is incorrect, then my mistake was starting the table with n = 0. 

If the correct way was to start the table with n = 1, then the fourth digit in the repeating group would be 1. 

 May 20, 2021
 #3
avatar+2205 
0

That's a wonderful explanation. :DD

But for 3^196 doesn't end with a 7, it ends with a 1. 

When you started counting, you started with a 0 so you have to add 1 spot to it.

 

=^._.^=

catmg  May 21, 2021

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