The digit in the unit place of the number (347)¹⁹² x (143)²⁰⁵ is :
9
1
7
3
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2023
For (347)¹⁹²: The unit digit of the base is 7. The cycle of unit digits for powers of 7 is 7, 9, 3, 1 (length 4). Divide the exponent 192 by the cycle length 4: 192 ÷ 4 = 48 with a remainder of 0. When the remainder is 0 (or the power is a multiple of the cycle length), the unit digit is the last digit in the cycle, which is 1 (corresponding to 7⁴). So, the unit digit of (347)¹⁹² is 1.
For (143)²⁰⁵: The unit digit of the base is 3. The cycle of unit digits for powers of 3 is 3, 9, 7, 1 (length 4). Divide the exponent 205 by the cycle length 4: 205 ÷ 4 = 51 with a remainder of 1. The unit digit is the first digit in the cycle, which is 3 (corresponding to 3¹). So, the unit digit of (143)²⁰⁵ is 3.
The unit digit of (347)¹⁹² x (143)²⁰⁵ is the unit digit of the product of the individual unit digits: unit digit of (1 * 3) = unit digit of 3, which is 3.
0: 0 (length 1)
1: 1 (length 1)
2: 2, 4, 8, 6 (length 4)
3: 3, 9, 7, 1 (length 4)
4: 4, 6 (length 2)
5: 5 (length 1)
6: 6 (length 1)
7: 7, 9, 3, 1 (length 4)
8: 8, 4, 2, 6 (length 4)
9: 9, 1 (length 2)
For a power ‘n’, divide ‘n’ by the cycle length. If the remainder is ‘r’ (r > 0), the unit digit is the r-th digit in the cycle. If the remainder is 0, the unit digit is the last digit in the cycle.