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How do you find anti-log of 3 digit number?
It is simply 10 to that power. Thus, antilog(231) = 10231. Could not be simpler.
Is a 3 digit number always greater than 2 digit number?
Yes, unless the 3 digit number is negative, or a decimal.
What is the greatest difference you can get when you subtract a 3 digit number from a 4 digit number?
Largest 4 digit number = 9999 Smallest 3 digit number = 111 9999 - 111 = 9888 Largest 4 digit number = 9999 Smallest 3 digit number = 111 9999 - 111 = 9888
When you divide a 3 digit number by a 1 digit number will the answer ever be a 1 digit number?
No.
Use a 3 digit number to form all two digit numbers?
By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3
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 53rd power?
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 3 to the 84th power?
7
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 200 power?
To find the units digit of 3 to the 200th power, we need to observe the pattern of units digits as we raise 3 to higher powers. The units digit of 3 to any power follows a repeating cycle: 3, 9, 7, 1. Since the cycle has a length of 4, we can divide 200 by 4 to find the remainder. 200 divided by 4 gives a remainder of 0, meaning the units digit of 3 to the 200th power is the last digit in the cycle, which is 1.
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
