Last digit of 39^22^23^24
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we take the base 39
there's a identity that (a^m)^n = a^m×n
so 39^(22×23×24) = 39^12144
we have to take the last digit so 9^4 = 6561
show the last digit is 1
hope it helps you
please mark it as brainliest
there's a identity that (a^m)^n = a^m×n
so 39^(22×23×24) = 39^12144
we have to take the last digit so 9^4 = 6561
show the last digit is 1
hope it helps you
please mark it as brainliest
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