The right digit of 4th power of any natural number

1,887

Solution 1

Using congruent modulo 10 $$x \equiv 0,1,2,3,...,8,9 \,\ (mod \,\ 10)$$ or, $$x \equiv 0,\pm1,\pm2,\pm3,\pm4,\pm5 \,\ (mod \,\ 10)$$ or, $$x^2 \equiv 0,1,4,9,16,25 \,\ (mod \,\ 10)$$ or, $$x^2 \equiv 0,1,4,9,6,5 \,\ (mod \,\ 10)$$ or, $$x^2 \equiv 0,\pm1,4,5,6 \,\ (mod \,\ 10)$$ or, $$x^4 \equiv 0,1,16,25,36 \,\ (mod \,\ 10)$$ or, $$x^4 \equiv 0,1,6,5,6 \,\ (mod \,\ 10)$$ or, $$x^4 \equiv 0,1,6,5 \,\ (mod \,\ 10)$$ $=>$ the rightmost digit of $x^4$ is either $0,1,6$ or $5$.

Solution 2

For the last digit of a product, you have to know only the last digits of the factors. Considering fourth powers, you only have 10 cases:

  • 1: $1^4 = 1$
  • 2: $2^4 = 16$
  • 3: $3^4 = 81$
  • 4: $4^4 = 256$
  • 5: $5^4 = 625$
  • 6: $6^4 = 1296$
  • 7: $7^4 = 2401$
  • 8: $8^4 = 4096$
  • 9: $9^4 = 6561$
  • 0: $0^4 = 0$

Looking at the last digits of these numbers, you see that only 0, 1, 5 and 6 appear. The last digits of these numbers have to be the last digit of any fourth power, for the reason stated above.

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Hamid Reza Ebrahimi
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Updated on July 22, 2022

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  • Hamid Reza Ebrahimi
    Hamid Reza Ebrahimi over 1 year

    Prove that the 4th power of any natural number has a right digit of 0,1,5 or 6.