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.
5500
Oh, what a happy little question! Let's paint a picture with numbers. When we raise 3 to any power, the last digit of the result follows a pattern: 3, 9, 7, 1, and then repeats. Since 444 is divisible by 4, the last digit of 3 to the power of 444 is 1. Let's add a touch of joy and say that the answer is a beautiful, calming 1.
answer is 3 sol: always divide power by 4 so here 333/4=83 and remainder is 1 then we take unit digit of the number which is 373 unit digit is 3 now 3^1 is 3 so units digit of 373^333 is 3
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.
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.
The units' digit of 222 to the power 666 is 4.
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.
5500
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 3. If you multiply 3 repeatedly by itself, you get the last digits:3, 9, 7, 1, 3, 9, 7, 1... In other words, every fourth power of 3 will end with the digit 1. Since any factorial starting at 4! has 4 as a factor, it follows that the last digit is 1.