What is the digit at the units place in the total integer value of 19^31 – 18^16 ?
Understand that in order to determine the units place, we do not need to calculate the full values of the exponents; we can get away with figuring out only the units digit. Let us consider each of the exponents separately - 19 ^ 31 - here instead of looking at 19 as a whole we can only take 9 and consider the units digit of 9. We have to check the pattern for the units digit occurence for powers of 9 - 9 ^ 1 = unit digit is 9 9 ^ 2 = unit digit is 1 9 ^ 3 = unit digit is 9 9 ^ 4 = unit digit is 1 so there is a repetitive pattern of either 9 or 1. As 31 is an odd number the unit digit will be = 9. Now let us consider 18 ^ 16. we will follow the same as above. 8 ^ 1 = unit digit is 88 ^ 2 = unit digit is 48 ^ 3 = unit digit is 28 ^ 4 = unit digit is 68 ^ 5 = unit digit is 88 ^ 6 = unit digit is 48 ^ 7 = unit digit is 28 ^ 8 = unit digit is 6here the repetitive pattern is of 8, 4, 2 or 6. As 16 is a multiple of 4 the units digit will be 6. Final answer is ----> 9 - 6 = 3.