0
Add your answer:
Earn +20 pts
What is the units digit of 7 to the 15 power?
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.
What is the units digit in 3 power 2011?
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.)
What is the units digit of the 5857th triangular number?
3
What is the units digit of the product when one hundred 7's are multiplied?
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.
What is the remainder when 3 to the power of 2008 divided by 5?
1ExplanationNote the following:31=332=933=2734=8135=24336=72937=218738=6561You will see that the units digit of the powers of three cycles through the following: 3, 9, 7, 1. 3 raised to a multiple of 4 has 1 as the units digit. Because 2008 is a multiple of 4, 32008 has 1 in the units digit (and is thus the remainder when that number is divided by 5).
What is the units digit of 7 to the 15 power?
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.
What is the units digit of 3 to the power of 19?
it is 3
What is the unit digit of 3 to the 60 power?
3 to a power divisible by 4 will have a units digit of 1.The powers of 3 are 3, 9, 27, 81 ... obviously, the next one will have a units digit of 1x3 or 3, the next one will have a units digit of 3x3 or 9, the next one will have a units digit of 7 (because 9x3 is 27), the next one will have a units digit of 1 (because 7x3 is 21), and then the cycle starts over with a units digit of 3 again.
What is the units digit of 2013 to the power of 2013?
Oh, dude, okay, so when you raise 2013 to the power of 2013, you're basically asking what the units digit of that massive number is. Well, lucky for you, you don't need to calculate the whole thing because the units digit of a number repeats in a pattern. So, the units digit of 2013 to the power of 2013 is 7. Cool, right?
What is the units digit of 3 to the 324 power?
1
What is the units digit of 3 to the 84th power?
7
How do you find the unit digit of 312 power 6?
Since neither the three hundred, nor the ten can contribute to the units digit in the answer, you look for a pattern in the units digit in the powers of 2n.20 = 121 = 222 = 423 = 824 = 2and after that , the pattern repeats, 4, 8, 2, 4, 8, 2, ...So if n (mod 3) = 1 the units digit is 2if n (mod 3) = 2 the units digit is 4and if n (mod 3) = 0 the units digit is 8where n (mod 3) is the remainder when n is divided by 3.312 is divisible by 3 [3+1+2=6 is divisible by 3] so 312 mod(3) =0 and so the units digit is 8.Since neither the three hundred, nor the ten can contribute to the units digit in the answer, you look for a pattern in the units digit in the powers of 2n.20 = 121 = 222 = 423 = 824 = 2and after that , the pattern repeats, 4, 8, 2, 4, 8, 2, ...So if n (mod 3) = 1 the units digit is 2if n (mod 3) = 2 the units digit is 4and if n (mod 3) = 0 the units digit is 8where n (mod 3) is the remainder when n is divided by 3.312 is divisible by 3 [3+1+2=6 is divisible by 3] so 312 mod(3) =0 and so the units digit is 8.Since neither the three hundred, nor the ten can contribute to the units digit in the answer, you look for a pattern in the units digit in the powers of 2n.20 = 121 = 222 = 423 = 824 = 2and after that , the pattern repeats, 4, 8, 2, 4, 8, 2, ...So if n (mod 3) = 1 the units digit is 2if n (mod 3) = 2 the units digit is 4and if n (mod 3) = 0 the units digit is 8where n (mod 3) is the remainder when n is divided by 3.312 is divisible by 3 [3+1+2=6 is divisible by 3] so 312 mod(3) =0 and so the units digit is 8.Since neither the three hundred, nor the ten can contribute to the units digit in the answer, you look for a pattern in the units digit in the powers of 2n.20 = 121 = 222 = 423 = 824 = 2and after that , the pattern repeats, 4, 8, 2, 4, 8, 2, ...So if n (mod 3) = 1 the units digit is 2if n (mod 3) = 2 the units digit is 4and if n (mod 3) = 0 the units digit is 8where n (mod 3) is the remainder when n is divided by 3.312 is divisible by 3 [3+1+2=6 is divisible by 3] so 312 mod(3) =0 and so the units digit is 8.
What is the units digit in 3 power 2011?
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.)
What is the last digit of 373 333?
To find the last digit of 373^333, we need to look for a pattern in the units digit of the powers of 3. The units digit of powers of 3 cycles every 4 powers: 3^1 = 3, 3^2 = 9, 3^3 = 7, 3^4 = 1, and then it repeats. Since 333 is one less than a multiple of 4, the units digit of 3^333 will be the third number in the cycle, which is 7. Therefore, the last digit of 373^333 is 7.
What is the units digit of the 5857th triangular number?
3
What is the units digit of 2143?
6308 6321
What is the remainder when 3 to the power of 2008 divided by 5?
1ExplanationNote the following:31=332=933=2734=8135=24336=72937=218738=6561You will see that the units digit of the powers of three cycles through the following: 3, 9, 7, 1. 3 raised to a multiple of 4 has 1 as the units digit. Because 2008 is a multiple of 4, 32008 has 1 in the units digit (and is thus the remainder when that number is divided by 5).
