Consider the following number :
n = [(6374)1793×(625)317×(313)49]
Which one of the following is the digit at the unit place of n ?
0
1
2
5
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CAPF – 2017
– Unit digit of (6374)^1793 is the unit digit of 4^1793. The pattern of unit digits for powers of 4 is 4, 6, 4, 6… The unit digit is 4 for odd exponents and 6 for even exponents. Since 1793 is odd, the unit digit of 4^1793 is 4.
– Unit digit of (625)^317 is the unit digit of 5^317. The unit digit of any positive integer power of 5 is always 5. So, the unit digit of 5^317 is 5.
– Unit digit of (313)^49 is the unit digit of 3^49. The pattern of unit digits for powers of 3 is 3, 9, 7, 1, 3, 9, 7, 1… The pattern repeats every 4 powers. We find the remainder of 49 divided by 4: 49 = 12 * 4 + 1. The remainder is 1. The unit digit is the same as the 1st power’s unit digit, which is 3. So, the unit digit of 3^49 is 3.
The unit digit of n is the unit digit of (Unit digit of 4^1793) * (Unit digit of 5^317) * (Unit digit of 3^49) = Unit digit of (4 * 5 * 3).
4 * 5 = 20. The unit digit of 20 is 0.
The unit digit of (20 * 3) is the unit digit of 60.
The unit digit is 0.