Find the last two digits (units and tens digit) in 1829^1829 and find the units digit in 23^7777.
See solution to the same problem at:
http://www.jiskha.com/display.cgi?id=1343870485
To find the last two digits of a number raised to a large power, such as 1829^1829 or 23^7777, we can use a concept called modular arithmetic.
Let's start with finding the last two digits of 1829^1829:
1. First, we need to break down the given number, 1829, into its units and tens digits. The units digit is 9, and the tens digit is 2.
2. Next, we analyze the patterns of the units digits when 1829 is raised to various powers. We want to find a pattern that repeats after a certain number of powers.
Let's calculate the units digits of 1829 raised to different powers:
- 1829^1 = 9
- 1829^2 = 8
- 1829^3 = 1
- 1829^4 = 9
- 1829^5 = 8
We notice that the units digit repeats every four powers: 9, 8, 1, 9, 8, 1, ... and so on.
3. Since the tens digit (2) does not affect the units digit, we only need to consider the units digit.
4. To find the units digit of 1829^1829, we divide the exponent (1829) by 4: 1829 ÷ 4 = 457 remainder 1.
5. The remainder (1) tells us which term in the repeating pattern we need to use. In this case, we need the first term, which is 9.
6. Therefore, the units digit of 1829^1829 is 9.
Now let's move on to finding the units digit of 23^7777:
1. We need to break down the given number, 23, into its units and tens digits. The units digit is 3, and the tens digit is 2.
2. Similar to the previous explanation, we observe the units digit pattern when 23 is raised to different powers:
- 23^1 = 3
- 23^2 = 9
- 23^3 = 7
- 23^4 = 1
- 23^5 = 3
We notice that the units digit follows a pattern that repeats every four powers: 3, 9, 7, 1, 3, 9, 7, ...
3. Again, since the tens digit (2) does not affect the units digit, we only need to consider the units digit.
4. To find the units digit of 23^7777, we divide the exponent (7777) by 4: 7777 ÷ 4 = 1944 remainder 1.
5. The remainder (1) indicates which term in the repeating pattern we need to use. In this case, we need the first term, which is 3.
6. Therefore, the units digit of 23^7777 is 3.
That's it! The last two digits of 1829^1829 are "29" (with 2 as the tens digit and 9 as the units digit). And the units digit of 23^7777 is "3".