Find the units digit of n given that \(mn = 21^6\) and m has a units digit of 7.
+128475 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
+6248 A bit of a different take on it that doesn't require you actually find the numbers.
\(\text{We are looking for }n \pmod{10}\\ m=10a+7\\ 21^6 = (2\cdot 10+1)^6 = \sum \limits_{k=0}^6 \dbinom{6}{k}(2\cdot 10)^k = 1 + 10\cdot (\text{some big number})\\ 21^6 \pmod{10} = 1\\ \)
\(mn \pmod{10} = (m \pmod{10})(n \pmod{10})\pmod{10} = 21^6 \pmod{10} = 1\\ m\pmod{10}=7\\ 7(n \pmod{10}) \pmod{10}=1\\ n\pmod{10} = 3 \)
