Let p = 2 2n + 2 + m and q = 2 4n - m (where n is even natural number). What should be the least value of m such that p as well as q is divisible by 5?
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Finding the Least Value of m for Divisibility by 5
The problem asks for the least value of the natural number m such that two expressions, \( p = 2^{2n + 2} + m \) and \( q = 2^{4n} - m \), are both divisible by 5, given that n is an even natural number.
For a number to be divisible by 5, it must leave a remainder of 0 when divided by 5. This can be expressed using modular arithmetic:
- \( p \equiv 0 \pmod{5} \)
- \( q \equiv 0 \pmod{5} \)
Understanding Powers of 2 Modulo 5
Let's examine the remainders when powers of 2 are divided by 5:
- \( 2^1 \equiv 2 \pmod{5} \)
- \( 2^2 = 4 \equiv 4 \pmod{5} \)
- \( 2^3 = 8 \equiv 3 \pmod{5} \)
- \( 2^4 = 16 \equiv 1 \pmod{5} \)
- \( 2^5 = 32 \equiv 2 \pmod{5} \)
The pattern of remainders is 2, 4, 3, 1, and it repeats every 4 powers. This means \( 2^k \pmod{5} \) depends on the value of \( k \pmod{4} \).
Analyzing Divisibility of p = 2<sup>2n + 2</sup> + m by 5
We are given that n is an even natural number. This means n can be written in the form \( n = 2k \) for some natural number \( k \ge 1 \). The exponent of 2 in the expression for p is \( 2n + 2 \). Substituting \( n = 2k \), we get the exponent \( 2(2k) + 2 = 4k + 2 \).
Now, let's find the remainder of the exponent when divided by 4:
\( (4k + 2) \pmod{4} \)
Since \( 4k \) is a multiple of 4, \( 4k \equiv 0 \pmod{4} \). Therefore,
\( 4k + 2 \equiv 0 + 2 \equiv 2 \pmod{4} \)
So, the exponent \( 2n+2 \) is always of the form \( 4k' + 2 \) for some integer k'. This means \( 2^{2n + 2} \equiv 2^2 \pmod{5} \).
\( 2^{2n + 2} \equiv 4 \pmod{5} \)
For p to be divisible by 5, we need \( p \equiv 0 \pmod{5} \):
\( 2^{2n + 2} + m \equiv 0 \pmod{5} \)
\( 4 + m \equiv 0 \pmod{5} \)
\( m \equiv -4 \pmod{5} \)
\( m \equiv 1 \pmod{5} \)
This tells us that m must leave a remainder of 1 when divided by 5.
Analyzing Divisibility of q = 2<sup>4n</sup> - m by 5
The exponent of 2 in the expression for q is \( 4n \). Since n is an even natural number, \( n = 2k \) for some natural number \( k \ge 1 \). The exponent becomes \( 4n = 4(2k) = 8k \).
Let's find the remainder of the exponent when divided by 4:
\( 8k \pmod{4} \)
Since \( 8k \) is always a multiple of 4 (as \( 8k = 4 \times 2k \)),
\( 8k \equiv 0 \pmod{4} \)
So, the exponent \( 4n \) is always a multiple of 4. This means \( 2^{4n} \equiv 2^0 \text{ or } 2^4 \equiv 1 \pmod{5} \).
\( 2^{4n} \equiv 1 \pmod{5} \)
For q to be divisible by 5, we need \( q \equiv 0 \pmod{5} \):
\( 2^{4n} - m \equiv 0 \pmod{5} \)
\( 1 - m \equiv 0 \pmod{5} \)
\( 1 \equiv m \pmod{5} \)
\( m \equiv 1 \pmod{5} \)
This confirms that m must leave a remainder of 1 when divided by 5.
Finding the Least Value of m
Both conditions require m to be a natural number such that \( m \equiv 1 \pmod{5} \). The natural numbers that satisfy this condition are those that leave a remainder of 1 when divided by 5. These are 1, 6, 11, 16, ...
We are looking for the least value of m from these possibilities. The least value is 1.
Let's check the given options:
| Option | Value of m | Is \( m \equiv 1 \pmod{5} \)? | Is it the least? |
|---|---|---|---|
| 1 | -1 | No (\( -1 \equiv 4 \pmod{5} \)) | N/A |
| 2 | 1 | Yes (\( 1 \equiv 1 \pmod{5} \)) | Yes, it is the smallest natural number satisfying the condition. |
| 3 | 4 | No (\( 4 \equiv 4 \pmod{5} \)) | N/A |
| 4 | 6 | Yes (\( 6 \equiv 1 \pmod{5} \)) | No, 1 is less than 6. |
The least value of m from the options that satisfies the required condition is 1.
Conclusion on Divisibility by 5 and Least m
For both \( p = 2^{2n + 2} + m \) and \( q = 2^{4n} - m \) to be divisible by 5, where n is an even natural number, m must satisfy \( m \equiv 1 \pmod{5} \). The least natural number m satisfying this condition is 1. Therefore, the least value of m is 1.
Revision Table: Key Concepts for Divisibility
| Concept | Description | Application in Problem |
|---|---|---|
| Divisibility by 5 | A number is divisible by 5 if its remainder when divided by 5 is 0 (\( x \equiv 0 \pmod{5} \)). | Used to set up congruences for p and q. |
| Modular Arithmetic | A system of arithmetic for integers where numbers "wrap around" upon reaching a certain value (the modulus). \( a \equiv b \pmod{m} \) means \( a - b \) is divisible by m. | Used to analyze the expressions p and q modulo 5. |
| Powers Modulo m | The sequence of remainders when powers of a base number are divided by a modulus m often follows a repeating pattern (a cycle). | Used to find \( 2^k \pmod{5} \) based on the pattern 2, 4, 3, 1. |
| Even Natural Number | A natural number that is divisible by 2 (2, 4, 6, ...). Represented as 2k where \( k \in \{1, 2, 3, ...\} \). | Used to determine the form of the exponents \( 2n+2 \) and \( 4n \). |
| Least Value | The smallest number in a set of possible values that satisfies a given condition. | Finding the minimum value of m from the set \( \{1, 6, 11, ...\} \). |
Additional Information: Properties of Modular Exponentiation
Modular exponentiation is a type of modular arithmetic performed with exponents. Calculating \( a^b \pmod{m} \) is often done efficiently by reducing the base modulo m at each step or by using properties of the cycle length of the powers.
- The sequence \( a^k \pmod{m} \) for \( k = 1, 2, 3, ... \) is always periodic. The length of this cycle is related to Euler's totient function, \( \phi(m) \).
- For powers of 2 modulo 5, the cycle length is 4, which divides \( \phi(5) = 5-1 = 4 \).
- To calculate \( 2^E \pmod{5} \), you only need to calculate \( E \pmod{4} \). If \( E \equiv r \pmod{4} \), then \( 2^E \equiv 2^r \pmod{5} \) (assuming \( r > 0 \) or E=0; special care needed if the modulus is not coprime to the base, but 5 is coprime to 2).
- In our problem, for \( 2^{2n+2} \pmod{5} \), the exponent is \( 2n+2 \). Since \( n=2k \), the exponent is \( 4k+2 \). \( 4k+2 \equiv 2 \pmod{4} \). So \( 2^{4k+2} \equiv 2^2 \equiv 4 \pmod{5} \).
- For \( 2^{4n} \pmod{5} \), the exponent is \( 4n \). Since \( n=2k \), the exponent is \( 8k \). \( 8k \equiv 0 \pmod{4} \). So \( 2^{8k} \equiv 2^0 \text{ or } 2^4 \equiv 1 \pmod{5} \).
These properties allow us to simplify the powers of 2 modulo 5 without needing to know the exact value of n, only its property of being an even natural number.
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