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Find the units digit of the number \(19^{ 2016 }+11^{ 2016 }-6^{ 2016 }\)

 Jul 31, 2020
 #1
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Instead of solving that (which I don't think anyone has the time for), we could just find the units digits of all those numbers and add and subtract them. 

How to we do that? Well, we know that 19^1 has a unit digit 9, and 19^2 has a unit digit 1, 19^3 has unit digit 9, 19^4 has unit digit 1... you see the patern? Every odd power has unit digit of 9, and even power gives unit digit of 1. Do the same thing with 11: 11^1 : 1, 11^2 : 1, 11^3 : 1... we see that every unit digit of a power of 11 gives 1. Now for 6: 6^1 : 6, 6^2 : 6, 6^3 : 6... same thing here, we see that the unit digit is 6 for every power of 6. 

So, now we do 1+1, which is 2. Borrow a 10 from the tenths place, 12-6 is 6. Therefore, the unit's digit is \(\boxed{6}\)

 Jul 31, 2020
 #2
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[19^2006 + 11^2006 - 6^2006] mod 10^10 =2,350,090,786

 Jul 31, 2020

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