The last digit,that is,the digit in the unit' | vedclass

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(A) To find the unit's digit of $(57)^{25}$,we look at the unit's digit of powers of $7$.$7^{1} = 7$$7^{2} = 49$ (unit's digit is $9$)$7^{3} = 343$ (u

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$6$

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(A) To find the unit's digit of $(57)^{25}$,we look at the unit's digit of powers of $7$.$7^{1} = 7$$7^{2} = 49$ (unit's digit is $9$)$7^{3} = 343$ (unit's digit is $3$)$7^{4} = 2401$ (unit's digit is $1$)The cycle of unit's digits is $(7, 9, 3, 1)$ with a period of $4$.For $(57)^{25}$,we divide the exponent $25$ by the period $4$: $25 = 4 	imes 6 + 1$.Since the remainder is $1$,the unit's digit of $(57)^{25}$ is the same as the unit's digit of $7^{1}$,which is $7$.Therefore,the unit's digit of $[(57)^{25} - 1]$ is $7 - 1 = 6$.

The last digit,that is,the digit in the unit's place of the number $[(57)^{25}-1]$ is

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