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Find the units digit of n given that \(mn = 21^6\) and m has a units digit of 7.

 Jan 7, 2019
 #1
avatar+101804 
+2

mn = 21^6

 

(7 * 3)^6  = 21^6

 

7^6  * 3^6  = 21^6

 

7^5 * 7 * 3^6  = 21^6

 

16807 * 7 *3^6 =  21^6

 

16807 * 5103 = 21^6

 

m = 16807

n = 5103

 

 

cool cool cool

 Jan 7, 2019
 #2
avatar+5225 
+3

A bit of a different take on it that doesn't require you actually find the numbers.

 

\(\text{We are looking for }n \pmod{10}\\ m=10a+7\\ 21^6 = (2\cdot 10+1)^6 = \sum \limits_{k=0}^6 \dbinom{6}{k}(2\cdot 10)^k = 1 + 10\cdot (\text{some big number})\\ 21^6 \pmod{10} = 1\\ \)

 

\(mn \pmod{10} = (m \pmod{10})(n \pmod{10})\pmod{10} = 21^6 \pmod{10} = 1\\ m\pmod{10}=7\\ 7(n \pmod{10}) \pmod{10}=1\\ n\pmod{10} = 3 \)

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 Jan 7, 2019

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