Question 213709
Last digit. Find the last digit in 3^9999.
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1st power last digit: 3
2nd power last digit: 9
3rd power last digit: 7
4rh power last digit: 1
5th power last digit: 3
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So there is a 5-step cycle
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9999/5 = 1999  with a remainder of 4
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So 3^(9999) = 3^(5*1999 + 4)
= (3^5)^1999 * 3^4
= 3*3^4
= 3^5
= 3
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Cheers,
Stan H.