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Bunuel
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We are asked for the unit's digit of 734^97 + 347^81

Let's remove the clutter and make it simpler to calculate
Units digit of 4^97 + 7^81 is the same as above

4^97
Even powers of 4 end with a 6 Odd powers of 4 end in 4
4^2 : 16 4^1: 4
4^4 : 256 4^3 : 64
4^6 : 4096 4^5 : 1024

So, 4^97 should end with a 4

7^81
7 has a power cyclicity of 4 which means the units digit repeats itself after every 4th power
7^1:7
7^2:49
7^3:343
7^4:2401
7^5:16807

accordingly, the 81st power should have the same units digit as the 1st i.e. 7

So , the addition should end with 4 + 7 = 11

Correct Option : B
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Unit digit of \(734^{97}\) = \(4^{97}\)

=> \(4^1\) = 4 [odd power will give 4 as unit digit] and \(4^2\) = 16 [even power will give 6 as unit digit]

=> \(734^{97}\) = 4 as unit digit

Unit digit of \(347^81\) = \(7^{81}\)

=> \(7^{4*20}\) * \(7^1\)

=> \(7^4\) * \(7^1\) = \(7^5\) [At \(5^{th}\) power, every number repeats itself at unit digit place].Therefore,

=>\(347^{81}\) = 7 as unit digit

Adding both 4 + 7 = 11.


Answer B
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