4 61 + 4 62 + 4 63 + 4 64 is divisible by
17
Understanding Divisibility of a Sum of Powers
The question asks us to determine which of the given options divides the sum of powers: \(4^{61} + 4^{62} + 4^{63} + 4^{64}\). To find the divisibility, we can simplify the expression by factoring out the lowest power of 4.
Step-by-Step Solution
Let the given expression be \(E\).
The expression is: \(E = 4^{61} + 4^{62} + 4^{63} + 4^{64}\).
We can see that \(4^{61}\) is the common factor in all the terms. Let's factor it out:
\(\qquad E = 4^{61} \times (1 + 4^{62-61} + 4^{63-61} + 4^{64-61})\)
\(\qquad E = 4^{61} \times (1 + 4^1 + 4^2 + 4^3)\)
Now, let's calculate the value of the expression inside the parenthesis:
\(\qquad 1 + 4^1 + 4^2 + 4^3 = 1 + 4 + (4 \times 4) + (4 \times 4 \times 4)\)
\(\qquad 1 + 4 + 16 + 64\)
Adding these numbers together:
\(\qquad 1 + 4 + 16 + 64 = 5 + 16 + 64 = 21 + 64 = 85\)
So, the expression simplifies to:
\(\qquad E = 4^{61} \times 85\)
Now, we need to determine which of the given options divides \(4^{61} \times 85\). A number is divisible by another number if the latter is a factor of the former. Let's look at the factors of 85.
We can find the prime factors of 85:
- 85 is not divisible by 2 (it's odd).
- The sum of digits is \(8+5=13\), which is not divisible by 3, so 85 is not divisible by 3.
- 85 ends in 5, so it is divisible by 5. \(85 \div 5 = 17\).
- 17 is a prime number.
So, the factors of 85 are 1, 5, 17, and 85. This means \(85 = 5 \times 17\).
The expression \(E\) can be written as:
\(\qquad E = 4^{61} \times 5 \times 17\)
Since the expression \(E\) is a product of \(4^{61}\), 5, and 17, it is divisible by any of these factors or their combinations. Looking at the options provided:
- Option 1: 7. Is 7 a factor of \(4^{61} \times 5 \times 17\)? No, 4 is \(2^2\), so \(4^{61}\) is \(2^{122}\). The prime factors are 2, 5, and 17. 7 is not among them.
- Option 2: 9. Is 9 a factor of \(4^{61} \times 5 \times 17\)? No. \(9 = 3^2\). The prime factors are 2, 5, and 17. 3 is not among them.
- Option 3: 11. Is 11 a factor of \(4^{61} \times 5 \times 17\)? No. 11 is not among the prime factors.
- Option 4: 17. Is 17 a factor of \(4^{61} \times 5 \times 17\)? Yes, 17 is explicitly a factor in the simplified expression.
Therefore, the sum \(4^{61} + 4^{62} + 4^{63} + 4^{64}\) is divisible by 17.
| Step | Calculation | Result |
|---|---|---|
| 1 | Factor out \(4^{61}\) | \(4^{61}(1 + 4^1 + 4^2 + 4^3)\) |
| 2 | Calculate sum inside parenthesis | \(1 + 4 + 16 + 64 = 85\) |
| 3 | Rewrite expression | \(4^{61} \times 85\) |
| 4 | Factorize 85 | \(85 = 5 \times 17\) |
| 5 | Final expression form | \(4^{61} \times 5 \times 17\) |
Conclusion
By factoring the expression and simplifying the sum inside the parenthesis, we found that the expression \(4^{61} + 4^{62} + 4^{63} + 4^{64}\) is equal to \(4^{61} \times 85\). Since \(85 = 5 \times 17\), the expression is \(4^{61} \times 5 \times 17\). This form clearly shows that the expression is divisible by 17.
Revision Table - Understanding Divisibility and Powers
| Concept | Description | Application in Problem |
|---|---|---|
| Factoring Common Terms | Pulling out a shared factor from an expression. Useful for simplification. | Factored out \(4^{61}\) from the sum. |
| Powers and Exponents | \(a^n\) means \(a\) multiplied by itself \(n\) times. \(a^{m+n} = a^m \times a^n\). | Used \(4^{61+k} = 4^{61} \times 4^k\) to factor terms. |
| Sum of Geometric Series (Partial) | \(1 + r + r^2 + \dots + r^n\). Here, \(r=4, n=3\). Sum is \(\frac{r^{n+1}-1}{r-1}\). | Calculated \(1+4+4^2+4^3\) manually, which is a sum of powers of 4. \(\frac{4^{3+1}-1}{4-1} = \frac{4^4-1}{3} = \frac{256-1}{3} = \frac{255}{3} = 85\). |
| Divisibility | A number 'a' is divisible by 'b' if 'b' is a factor of 'a'. | Checked if 17 is a factor of the simplified expression \(4^{61} \times 85\). |
| Prime Factorization | Writing a number as a product of its prime factors. | Factorized \(85 = 5 \times 17\) to identify divisors. |
Additional Information - Properties of Exponents and Divisibility
Understanding how exponents work and the rules of divisibility is crucial for solving such problems.
- Exponent Rule: When multiplying powers with the same base, you add the exponents: \(a^m \times a^n = a^{m+n}\). Conversely, \(a^{m+n} = a^m \times a^n\). This rule was key in factoring \(4^{61}\). For example, \(4^{62} = 4^{61+1} = 4^{61} \times 4^1\).
- Factoring: Factoring out a common term helps simplify complex expressions and reveal their structure, making it easier to identify divisors.
- Divisibility by Factors: If a number \(N\) can be written as a product of integers, say \(N = a \times b \times c\), then \(N\) is divisible by \(a\), by \(b\), by \(c\), and by any product of these factors (like \(a \times b\), \(b \times c\), \(a \times b \times c\)). In our case, the expression is \(4^{61} \times 5 \times 17\), so it is divisible by 17.
- Divisibility Tests: Knowing basic divisibility tests (by 2, 3, 5, etc.) can help factor numbers like 85 quickly.
This type of problem is common in mathematics and helps build skills in algebraic manipulation and number theory.
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