Find the digit at the unit’s place of 10759×31387×438129×39999

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Find the digit at the unit’s place of ${(107)}^{59} \times {(313)}^{87} \times {(438)}^{129} \times {(399)}^{99}$

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answer is A.

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Detailed Solution

We are given the expression

{(107)}^{59} \times {(313)}^{87} \times {(438)}^{129} \times {(399)}^{99}

Consider the last digits of 107, 313, 438, and 399.
So, the last digits are 7, 3, 8, and 9.
We know that the last digit of

{(107)}^{59} \times {(313)}^{87} \times {(438)}^{129} \times {(399)}^{99}

is same as the last digit of

{(7)}^{59} \times {(3)}^{87} \times {(8)}^{129} \times {(9)}^{99} . . . . . . . . . (I)

Also, the cyclicity of 7, 3, 8, and 9 are 4, 4, 4, and 2 respectively.
Consider the cyclicity charts of 7, 3, 8, and 9
71=7, 73= 343, 72= 49. 74= 2401
Thus, digit in the unit’s place of 74=1
Consider the power 59 of 7. We can write 59=4×14+3

\Rightarrow {(7)}^{59} = {(7)}^{4 \times 14 + 3} = {(7^{4})}^{14} . 73 l

Therefore, last digit of (7)59 = last digit of

{(7^{4})}^{14}

× last digit of 73
As the last digit of 74 is 1, the last digit of any power of 74 is 1.
Also, as seen from the cyclicity chart, last digit of 73 is 3……..(1)
Similarly, with the help of cyclicity, we get the last digits of 34,84,92 as follows:
Last digit of 34 is 1.
Last digit of 84 is 6.
Last digit of 92 is 1.
Also,
(3)87=(3)4×21+4=(34)21.33
(8)129=(8)4×32+1=(84)14.81
(9)99=(9)2×49+1=(92)49.91
Thus, last digit of (3)87=7..........(2)
Last digit of (8)129=8..........(3)
Last digit of (9)99=9..........(4)
Thus, from (I), (1), (2), (3), and (4), we have