What number is obtained by adding the units digits of 734^97 and 347^8

Units digit of unit 4 to the power of antything:
4 odd
6 even

Units digit is 4

Units digit of unit 7 to the power of anything
7
9
3
1
----> cycle formed. So every four numbers, the fifth number the cycle starts again. 80/4 means 20 cycles, 81 a new cycle starts so 7 as unit
7
9
3

4+7=11 Answer is B beyond any doubt.

What number is obtained by adding the units digits of 734^97 and 347^81?

A. 10
B. 11
C. 12
D. 13
E. 14

Units digit of 734^97 is 4. (4 raised to the power odd integer).
Units digit of 347^81 is 7. (Powers of 7 end in 7,9,3 or 1. Since 81 is in the form 4n+1; the UD is 7).

therefore 7+4 = 11 Answer B.

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

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