1
The units digit of 20132013 is the same as the units digit of 32013. The units digit of 34 = units digit of 81 = 1 So units digit of 32013 = 32012+1 = 34*503+1 = 34*503 *31 = 1503*3 = 3
To find the units digit of (27^{27}), we can look at the units digit of (27), which is (7). We then need to find the units digit of (7^{27}). The units digits of the powers of (7) cycle every four terms: (7^1 = 7), (7^2 = 49) (units digit (9)), (7^3 = 343) (units digit (3)), and (7^4 = 2401) (units digit (1)). Since (27 \mod 4 = 3), the units digit of (7^{27}) is the same as that of (7^3), which is (3). Thus, the units digit of (27^{27}) is (3).
7
To solve a question like this, one looks at patterns of powers. For example: 31 = 3 32 = 9 33 = 27 34 = 81 Hence 34 = 1 and 38 = 1 and 312 = 1 and 316 = 1 and so on ............ now in this same sequence 3200 = 1 hence unit digit is 1.
Its positional place value is seven ones or units = 7
The units digit of 20132013 is the same as the units digit of 32013. The units digit of 34 = units digit of 81 = 1 So units digit of 32013 = 32012+1 = 34*503+1 = 34*503 *31 = 1503*3 = 3
Power 2: units digit 9. Multiply by 49 again to get power 4: units digit 1. So every 4th power gives units digit 1. So 16th power has units digit 1, so the previous power, the 15th must have units digit 3.
To find the units digit of (27^{27}), we can look at the units digit of (27), which is (7). We then need to find the units digit of (7^{27}). The units digits of the powers of (7) cycle every four terms: (7^1 = 7), (7^2 = 49) (units digit (9)), (7^3 = 343) (units digit (3)), and (7^4 = 2401) (units digit (1)). Since (27 \mod 4 = 3), the units digit of (7^{27}) is the same as that of (7^3), which is (3). Thus, the units digit of (27^{27}) is (3).
7
When multiplying numbers with the same units digit, the units digit of the product is determined by the units digit of the base number raised to the power of the number of times it is being multiplied. In this case, since 7 is being multiplied 100 times, the units digit of the product will be the same as the units digit of 7^100. The units digit of 7^100 can be found by looking for a pattern in the units digits of powers of 7: 7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, and so on. The pattern repeats every 4 powers, so the units digit of 7^100 will be the same as 7^4, which is 1. Therefore, the units digit of the product when one hundred 7's are multiplied is 1.
To solve a question like this, one looks at patterns of powers. For example: 31 = 3 32 = 9 33 = 27 34 = 81 Hence 34 = 1 and 38 = 1 and 312 = 1 and 316 = 1 and so on ............ now in this same sequence 3200 = 1 hence unit digit is 1.
The digit in the units column of the number 7157 is 7.
Well, isn't that a happy little math problem! When we look at the unit digit of powers of numbers, we focus on the cyclical pattern they follow. The unit digit of 3 raised to any power follows a pattern: 3, 9, 7, 1, and then repeats. So, to find the unit digit of 3 to the power of 34 factorial, we look for the remainder when 34 factorial is divided by 4, which is 2. Therefore, the unit digit of 3 to the power of 34 factorial is 9.
To find the units' digit of 3 to the power of 333, we need to look for a pattern. The units' digit of powers of 3 cycles in a pattern: 3, 9, 7, 1. Since 333 divided by 4 leaves a remainder of 1, the units' digit of 3 to the power of 333 will be the first digit in the pattern, which is 3.
7 7 7 7
I guess you mean what's the units digit of 32011. It is 7. To work this out, see how the units digit of 3n changes; it goes: 3, 9, 7, 1, 3, 9, 7, 1, ... (only the first 8 powers are shown) repeating the same sequence of 4 digits. So if we find the remainder of 2011 divided by 4, it will tell us which of the four numbers (3, 9, 7, 1) will be the units digit of 32011: 2011 ÷ 4 ⇒ remainder 3, so the 3rd digit is the required digit: 7. (If there had been no remainder, then the 4th digit, namely 1, would have been the required value.)
Its positional place value is seven ones or units = 7