Table of Contents
How do you find the last two digits of an exponent?
Case 2: Units digit in x is 3, 7 or 9
- When x ends in 9.
- Raise the base by 2 and divide the exponent by 2 =>
- Number ending in 9 raised to 2 ends in 1 =>
- When x ends in 3.
- Raise the base by 4 and divide the exponent by 4 =>
- Number ending in 3 raised to 4 ends in 1 =>
- When x ends in 7.
What is the last digit of 2¹⁰⁰?
The last four digits are 5376, the last digit is 6. Except for 2^0, two to any larger power must be even, therefore only {2,4,6,8}, not even zero, are the only possible last digit.
What is the last digit of 2017 to the power 2017?
Rather, the remainder is 1 (i.e. 2017 = 4*504 + 1). So 7^2017 = (7^4)^504 * 7^1 = 7. So the last digit is 7.
What is the last digit of 2017?
What is the last digit of 333?
So, when 333 is divided by 4 the remainder is 1, so 373 raised to the power 1 is 373, hence the last digit of the result is 373 raise to the power that is 3.
What is the last digit of 6¹⁰⁰?
6 to any power ends with 6.
How do you find the last 2 digits of an integer?
Finding the last digit of a positive integer is the same as finding the remainder of that number when divided by 10 10. In general, the last digit of a power in base n n. For decimal numbers, we compute \\bmod~ {10} mod 10 . Finding the last 2 digits of an integer amounts to computing it mod
What is the unit digit of 2014 to the power of 2012?
The base is 2014 and hence its units digit is 4. Therefore, l = 4. The exponent is 2012, which is divisible by 4. Since l is even and the exponent is divisible by 4, we have the units digit of 2014 to the power of 2012 is 6. The base is a single digit 3. Therefore, l = 3.
What is the last digit of a 4 digit number pattern?
The last digit repeats in a pattern that is 4 digits long: 7,9,3,1 7,9,3,1.
How do you find the last 2 digits of a power?
In general, the last digit of a power in base n n n is its remainder upon division by n n n. For decimal numbers, we compute m o d 10 \\bmod~{10} m o d 1 0 . Finding the last 2 digits of an integer amounts to computing it mod 100 , 100, 1 0 0 , and finding the last n {n} n digits amounts to computation m o d 1 0 n \\bmod~10^{n} m o d 1 0 n .