The last digit is 4.
It is the last digit of 34= 81. Therefore it is 1.It is the last digit of 34= 81. Therefore it is 1.It is the last digit of 34= 81. Therefore it is 1.It is the last digit of 34= 81. Therefore it is 1.
6. All even powers of 4 end in 6
I assume you mean the last digit. The idea is to find a pattern: 8 to the power 1 ends with 8 8 to the power 2 ends with 4 8 to the power 3 ends with 2 (to figure this out, I just multiplied 4 x 8) 8 to the power 4 ends with 6 (once again, multiply the last digit of the previous power x 8) 8 to the power 5 ends with 8 8 to the power 6 ends with 4 ... As you can see, it all repeats, with a period of 4. Therefore, any exponent that is a multiple of 4 will end with 6. Find the nearest multiple of 4 that is smaller than 2015, then continue calculating - always with the last digit.
The digit with which a multiple of 4 ends depends on the last digit of the other factor. If the last digit is a zero, the product ends with zero; if the last digit is a 1, the product ends with 4; etc. The only options for the last digit of the product are 0, 2, 4, 6, 8.
The last digit is 4.
4
To find the last digit of a number raised to a power, we can use the concept of modular arithmetic. The last digit of 333 to the power of 444 can be determined by finding the remainder when 333 is divided by 10, which is 3. Since the last digit of 333 is 3, we need to find the remainder of 444 divided by 4, which is 0. Therefore, the last digit of 333 to the power of 444 is the same as the last digit of 3 to the power of 4, which is 1.
The units' digit of 222 to the power 666 is 4.
It is the last digit of 34= 81. Therefore it is 1.It is the last digit of 34= 81. Therefore it is 1.It is the last digit of 34= 81. Therefore it is 1.It is the last digit of 34= 81. Therefore it is 1.
6. All even powers of 4 end in 6
4 x 4 = 16. If you multiply the last digit by 4: 6 x 4 = 24 (ends with 4). Continue multiplying by 4, and you see that the last digit will always alternate between 4 and 6 - 4 for the odd powers, and 6 for the even powers. So, in this case, the answer is 4.
I assume you mean the last digit. The idea is to find a pattern: 8 to the power 1 ends with 8 8 to the power 2 ends with 4 8 to the power 3 ends with 2 (to figure this out, I just multiplied 4 x 8) 8 to the power 4 ends with 6 (once again, multiply the last digit of the previous power x 8) 8 to the power 5 ends with 8 8 to the power 6 ends with 4 ... As you can see, it all repeats, with a period of 4. Therefore, any exponent that is a multiple of 4 will end with 6. Find the nearest multiple of 4 that is smaller than 2015, then continue calculating - always with the last digit.
The digit with which a multiple of 4 ends depends on the last digit of the other factor. If the last digit is a zero, the product ends with zero; if the last digit is a 1, the product ends with 4; etc. The only options for the last digit of the product are 0, 2, 4, 6, 8.
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.
A number is a multiple of 4 if the last 2 digits are a multiple of 4 The 10s digit is even and the last digit is 0, 4 or 8 The 10s digit is odd and the last digit is 2 or 6 A number is a multiple of 8 if the last 3 digits are a multiple of 8 The 100s digit is even and the last 2 digits are a multiple of 8 The 100s digit is odd and the last 2 digits are 4 times an odd number
If the last two digits are divisible by 4. This is equivalent to: Last digit = 0, 4, 8 and the digit before (in tens place) is even or Last digit = 2 or 6 and the digit before (in tens place) is odd.