What is the digit that will come at the unit’s place in the result of (423) 423 × (1237) 28 ?
7
Finding the Unit Digit of a Product of Powers
This question asks for the unit digit of the result obtained by multiplying two numbers raised to certain powers: &\((423)^{423} \times (1237)^{28}&\). To find the unit digit of a product, we only need to find the unit digit of each factor and then find the unit digit of their product.
The unit digit of a number raised to a power depends only on the unit digit of the base and the exponent. The pattern of unit digits for powers of any number repeats in a cycle.
Step 1: Find the Unit Digit of \((423)^{423}\)
The unit digit of the base number 423 is 3. We need to look at the pattern of the unit digits of powers of 3.
- \(3^1 = 3\) (Unit digit is 3)
- \(3^2 = 9\) (Unit digit is 9)
- \(3^3 = 27\) (Unit digit is 7)
- \(3^4 = 81\) (Unit digit is 1)
- \(3^5 = 243\) (Unit digit is 3)
The pattern of unit digits for powers of 3 is 3, 9, 7, 1. This cycle has a length of 4.
To find the unit digit of \((423)^{423}\), we look at the power, which is 423. We divide the power by the length of the cycle (4) and look at the remainder.
Divide 423 by 4:
\begin{equation*} 423 \div 4 \end{equation*}
\begin{equation*} 423 = 4 \times 105 + 3 \end{equation*}
The remainder is 3.
If the remainder is 1, the unit digit is the 1st in the cycle (3). If the remainder is 2, it's the 2nd (9). If the remainder is 3, it's the 3rd (7). If the remainder is 0 (for powers that are multiples of 4), the unit digit is the last in the cycle (1).
Since the remainder is 3, the unit digit of \((423)^{423}\) is the 3rd digit in the cycle (3, 9, 7, 1), which is 7.
Step 2: Find the Unit Digit of \((1237)^{28}\)
The unit digit of the base number 1237 is 7. We need to look at the pattern of the unit digits of powers of 7.
- \(7^1 = 7\) (Unit digit is 7)
- \(7^2 = 49\) (Unit digit is 9)
- \(7^3 = 343\) (Unit digit is 3)
- \(7^4 = 2401\) (Unit digit is 1)
- \(7^5 = 16807\) (Unit digit is 7)
The pattern of unit digits for powers of 7 is 7, 9, 3, 1. This cycle also has a length of 4.
To find the unit digit of \((1237)^{28}\), we look at the power, which is 28. We divide the power by the length of the cycle (4) and look at the remainder.
Divide 28 by 4:
\begin{equation*} 28 \div 4 \end{equation*}
\begin{equation*} 28 = 4 \times 7 + 0 \end{equation*}
The remainder is 0.
When the remainder is 0 (meaning the power is a multiple of 4), the unit digit is the last digit in the cycle (7, 9, 3, 1), which is 1.
So, the unit digit of \((1237)^{28}\) is 1.
Step 3: Find the Unit Digit of the Product
The unit digit of the product \((423)^{423} \times (1237)^{28}\) is the unit digit of the product of their individual unit digits.
- Unit digit of \((423)^{423}\) is 7.
- Unit digit of \((1237)^{28}\) is 1.
The unit digit of the product is the unit digit of \(7 \times 1\).
\begin{equation*} 7 \times 1 = 7 \end{equation*}
The unit digit of the result is 7.
Unit Digit Cyclicity Summary
The pattern of unit digits for powers of a number is periodic. The length of the cycle (cyclicity) is usually 4 for most digits (2, 3, 7, 8). Digits 0, 1, 5, 6 have a cyclicity of 1. Digits 4 and 9 have a cyclicity of 2.
| Unit Digit of Base | Pattern of Unit Digits of Powers | Cyclicity |
|---|---|---|
| 0 | 0 | 1 |
| 1 | 1 | 1 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 4 | 4, 6 | 2 |
| 5 | 5 | 1 |
| 6 | 6 | 1 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
| 9 | 9, 1 | 2 |
Revision Table: Key Concepts for Unit Digits
| Concept | Explanation | How it applies here |
|---|---|---|
| Unit Digit of a Power | Only depends on the unit digit of the base and the exponent. | We focused on the unit digits 3 and 7. |
| Cyclicity of Unit Digits | The pattern of unit digits for powers repeats in a cycle. | Digits 3 and 7 have a cycle of length 4. |
| Using Remainders | Divide the exponent by the cyclicity. The remainder tells you the position in the cycle. Remainder 0 means the last digit in the cycle. | For \((423)^{423}\): \(423 \div 4\), Remainder = 3. Unit digit is 3rd in cycle (7). For \((1237)^{28}\): \(28 \div 4\), Remainder = 0. Unit digit is 4th (last) in cycle (1). |
| Unit Digit of a Product | The unit digit of a product is the unit digit of the product of the unit digits of the factors. | Unit digit of \((423)^{423} \times (1237)^{28}\) is the unit digit of \(7 \times 1\), which is 7. |
Additional Information: Understanding Cyclicity
Cyclicity in finding unit digits is a fundamental concept. It arises because the unit digit of a product depends only on the unit digits of the numbers being multiplied. When you repeatedly multiply a number by itself, the unit digit sequence eventually repeats.
For example, with the unit digit 3:
- \(3^1 \rightarrow\) unit digit 3
- \(3^2 \rightarrow 3 \times 3 = 9 \rightarrow\) unit digit 9
- \(3^3 \rightarrow 9 \times 3 = 27 \rightarrow\) unit digit 7
- \(3^4 \rightarrow 7 \times 3 = 21 \rightarrow\) unit digit 1
- \(3^5 \rightarrow 1 \times 3 = 3 \rightarrow\) unit digit 3 (The pattern restarts)
This cycle (3, 9, 7, 1) of length 4 will continue for all higher powers of any number ending in 3. To find the unit digit for a specific power, say \(3^N\), you just need to see where \(N\) falls in this cycle, which is determined by the remainder when \(N\) is divided by 4.
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