What is the last digit of the sum S = 927 + 279 ?
6
To find the last digit of a sum like \(S = 9^{27} + 27^9\), we need to determine the last digit of each term separately and then find the last digit of their sum.
Finding the Last Digit of Powers
The last digit of a power depends only on the last digit of the base and the exponent. We observe the pattern of the last digits as the exponent increases.
Finding the Last Digit of \(9^{27}\)
Let's look at the last digits of the first few powers of 9:
- \(9^1\): The last digit is 9.
- \(9^2 = 81\): The last digit is 1.
- \(9^3 = 729\): The last digit is 9.
- \(9^4 = 6561\): The last digit is 1.
The pattern of the last digits of powers of 9 is 9, 1, 9, 1, ... This pattern repeats every 2 powers. We can see that the last digit is 9 when the exponent is odd, and 1 when the exponent is even.
For \(9^{27}\), the exponent is 27, which is an odd number.
Therefore, the last digit of \(9^{27}\) is 9.
Finding the Last Digit of \(27^9\)
The last digit of \(27^9\) is determined by the last digit of the base, which is 7. So, we need to find the last digit of \(7^9\). Let's look at the last digits of the first few powers of 7:
- \(7^1\): The last digit is 7.
- \(7^2 = 49\): The last digit is 9.
- \(7^3 = 343\): The last digit is 3.
- \(7^4 = 2401\): The last digit is 1.
- \(7^5 = 16807\): The last digit is 7.
The pattern of the last digits of powers of 7 is 7, 9, 3, 1, 7, 9, 3, 1, ... This pattern repeats every 4 powers. The cycle of the last digits is (7, 9, 3, 1).
To find the last digit of \(7^9\), we need to find where the 9th term falls in this cycle of length 4. We do this by dividing the exponent 9 by the cycle length 4 and looking at the remainder.
Exponent = 9
Cycle length = 4
\(9 \div 4 = 2\) with a remainder of 1.
The remainder is 1. This means the last digit is the same as the 1st digit in the cycle (7, 9, 3, 1), which is 7.
Therefore, the last digit of \(27^9\) (which is the same as \(7^9\)) is 7.
Finding the Last Digit of the Sum \(S = 9^{27} + 27^9\)
To find the last digit of the sum \(S = 9^{27} + 27^9\), we add the last digits of the individual terms and find the last digit of that sum.
- Last digit of \(9^{27}\) is 9.
- Last digit of \(27^9\) is 7.
Sum of the last digits = \(9 + 7 = 16\).
The last digit of 16 is 6.
So, the last digit of the sum \(S = 9^{27} + 27^9\) is 6.
Revision Table: Last Digit Patterns
| Base's Last Digit | Pattern of Last Digits | Cycle Length |
|---|---|---|
| 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 |
Additional Information on Cyclicity of Last Digits
The property we used to find the last digit of powers is called cyclicity. The last digits of powers of any integer greater than 1 follow a repeating pattern. The length of this repeating pattern, or cycle length, depends on the last digit of the base number. For bases ending in 0, 1, 5, or 6, the cycle length is 1. For bases ending in 4 or 9, the cycle length is 2. For bases ending in 2, 3, 7, or 8, the cycle length is 4.
To find the last digit of a number \(X^N\), where \(X\) ends in \(d\) (one of 2, 3, 7, or 8), you find the remainder of \(N\) when divided by 4. If the remainder is \(r\), the last digit is the \(r\)-th digit in the cycle for base \(d\). If the remainder is 0, the last digit is the last digit in the cycle (which corresponds to the 4th power).
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