Re: Units digit of (233)^{23}(727
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27 Jun 2025, 04:00
To find the units digit of a product, we only need to consider the units digit of each number in the product and the pattern of units digits when raised to powers.
We need to find the units digit of $\((233)^{23}(727)^{22}(929)^3\)$.
This is equivalent to finding the units digit of $\((3)^{23}(7)^{22}(9)^3\)$.
Let's find the cycle of the units digits for powers of 3,7 , and 9 .
1. Units digit of $\((3)^{23}\)$ :
- $\(3^1=3\)$
- $\(3^2=9\)$
- $\(3^3=27 \Longrightarrow 7\)$
- $\(3^4=81 \Longrightarrow 1\)$
- $\(3^5=243 \Longrightarrow 3\)$
The cycle of units digits for powers of 3 is ( $3,9,7,1$ ), which has a length of 4. To find the units digit of $\(3^{23}\)$, divide the exponent 23 by the cycle length 4 : $\(23 \div 4=5\)$ with a remainder of 3 .
So, the units digit of $3^{23}$ is the same as the 3rd digit in the cycle, which is 7 .
2. Units digit of $\((7)^{22}\)$ :
- $\(7^1=7\)$
- $\(7^2=49 \Longrightarrow 9\)$
- $\(7^3=343 \Longrightarrow 3\)$
- $\(7^4=2401 \Longrightarrow 1\)$
- $\(7^5=16807 \Longrightarrow 7\)$
The cycle of units digits for powers of 7 is ( $7,9,3,1$ ), which has a length of 4. To find the units digit of $\(7^{22}\)$, divide the exponent 22 by the cycle length 4 : $\(22 \div 4=5\)$ with a remainder of 2 .
So, the units digit of $\(7^{22}\)$ is the same as the 2 nd digit in the cycle, which is 9 .
3. Units digit of $\((9)^3\)$ :
- $\(9^1=9\)$
- $\(9^2=81 \Longrightarrow 1\)$
- $\(9^3=729 \Longrightarrow 9\)$
The cycle of units digits for powers of 9 is $(9,1)$, which has a length of 2 .
To find the units digit of $9^3$, divide the exponent 3 by the cycle length 2 :
$\(3 \div 2=1\)$ with a remainder of 1 .
So, the units digit of $9^3$ is the same as the 1st digit in the cycle, which is 9.
4. Find the units digit of the product:
Multiply the units digits we found:
Units digit of $\((3)^{23} \times(7)^{22} \times(9)^3\)$
Units digit of $\(7 \times 9 \times 9\)$
First, $\(7 \times 9=63\)$. The units digit is 3 .
Then, multiply this units digit by the last 9 :
Units digit of $\(3 \times 9=27\)$. The units digit is 7 .
So, Quantity A is 7.
Compare Quantity A and Quantity B:
Quantity A: 7
Quantity B: 1
Therefore, Quantity A is greater than Quantity B.
The final answer is Quantity A is greater.