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# Find the last digit of `7^121 -3^402`

bayleef999 | Student | eNoter

Posted June 19, 2013 at 4:41 AM via web

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Find the last digit of `7^121 -3^402`

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 19, 2013 at 5:15 AM (Answer #1)

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The last digit of `7^121 - 3^402` has to be determined.

For 7: `7^0 = 1` , `7^1 = 7` , `7^2 = 49` , `7^3 = 343` , `7^4 = 2401` .

The last digit of the power of 7 is the same for values of the exponent that differ by 4.

`7^121 = 7^(4*30 + 1)` . This gives the last digit of `7^(4*30 + 1)` as 7

For 3: `3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81`

The last digit of the power of 3 is the same for values of the exponent that differ by 4.

`3^402 = 3^(2 + 4*100)` . This gives the last digit of `3^402` as 9.

`3^402 = 9^201` , this is greater than `7^121` . The result of `7^121 - 3^402` is a negative number. The last digit of the negative number is 9 - 7 = 2

The last digit of `7^121 - 3^402` is 2.

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