Comprehension
Functions \( g \) and \( h \) are defined on \( n \) constants, \( a_0, a_1, a_2, a_3, \ldots, a_{n-1} \), as follows:
\[ g(a_p, a_q) = \begin{cases} a_{|p - q|}, & \text{if } |p - q| \leq (n - 4) \\ a_{n - |p - q|}, & \text{if } |p - q| > (n - 4) \end{cases} \]
\[ h(a_p, a_q) = a_k, \quad \text{where } k \text{ is the remainder when } (p + q) \text{ is divided by } n. \]
\[ g(a_p, a_q) = \begin{cases} a_{|p - q|}, & \text{if } |p - q| \leq (n - 4) \\ a_{n - |p - q|}, & \text{if } |p - q| > (n - 4) \end{cases} \]
\[ h(a_p, a_q) = a_k, \quad \text{where } k \text{ is the remainder when } (p + q) \text{ is divided by } n. \]
Question: 1
If \( n = 10 \), find the value of \( g(g(a_2, a_8), g(a_1, a_7)) \).
If \( n = 10 \), find the value of \( g(g(a_2, a_8), g(a_1, a_7)) \).
Show Hint
Always compute absolute difference and compare with \( n - 4 \) to decide which case of function \( g \) applies.
Updated On: Jul 28, 2025
- \( a_9 \)
- \( a_7 \)
- \( a_2 \)
- \( a_0 \)
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The Correct Option is D
Solution and Explanation
We are given the function \( g(a_p, a_q) \) defined as follows:
If \( |p - q| \leq (n - 4) \), then \( g(a_p, a_q) = a_{|p - q|} \)
Otherwise, \( g(a_p, a_q) = a_{n - |p - q|} \)
Step 1: Compute \( g(a_2, a_8) \):
\(|2 - 8| = 6 \), and since \( 6 \leq 6 \), use first case:
\(\Rightarrow g(a_2, a_8) = a_6 \)
Step 2: Compute \( g(a_1, a_7) \):
\(|1 - 7| = 6 \), and \( 6 \leq 6 \), so:
\(\Rightarrow g(a_1, a_7) = a_6 \)
Step 3: Compute \( g(a_6, a_6) \):
\(|6 - 6| = 0 \), and \( 0 \leq 6 \), so:
\(\Rightarrow g(a_6, a_6) = a_0 \)
Final Answer: \(\boxed{a_0}\)
If \( |p - q| \leq (n - 4) \), then \( g(a_p, a_q) = a_{|p - q|} \)
Otherwise, \( g(a_p, a_q) = a_{n - |p - q|} \)
Step 1: Compute \( g(a_2, a_8) \):
\(|2 - 8| = 6 \), and since \( 6 \leq 6 \), use first case:
\(\Rightarrow g(a_2, a_8) = a_6 \)
Step 2: Compute \( g(a_1, a_7) \):
\(|1 - 7| = 6 \), and \( 6 \leq 6 \), so:
\(\Rightarrow g(a_1, a_7) = a_6 \)
Step 3: Compute \( g(a_6, a_6) \):
\(|6 - 6| = 0 \), and \( 0 \leq 6 \), so:
\(\Rightarrow g(a_6, a_6) = a_0 \)
Final Answer: \(\boxed{a_0}\)
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Question: 2
If \( h(a_k, a_m) = a_m \) for all \( m \), where \( 1 \leq m<n \) and \( 0 \leq k<n \), and \( m \) is a natural number, find \( k \).
If \( h(a_k, a_m) = a_m \) for all \( m \), where \( 1 \leq m<n \) and \( 0 \leq k<n \), and \( m \) is a natural number, find \( k \).
Show Hint
Set up the modulo equation from the functional definition and simplify to find valid values of \( k \).
Updated On: Jul 28, 2025
- \( 0 \)
- \( 1 \)
- \( n - 1 \)
- \( n - 2 \)
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The Correct Option is A
Solution and Explanation
We are given that the function \( h(a_p, a_q) = a_k \), where \( k = (p + q) \mod n \).
Also, \( h(a_k, a_m) = a_m \) for all \( m \Rightarrow (k + m) \mod n = m \)
Step 1: Solve the congruence
\( (k + m) \mod n = m \Rightarrow k \mod n = 0 \Rightarrow k = 0 \)
Verification: For \( k = 0 \),
\( h(a_0, a_m) = a_{(0 + m) \mod n} = a_m \) which satisfies the condition for all \( m \).
Final Answer: \(\boxed{0}\)
Also, \( h(a_k, a_m) = a_m \) for all \( m \Rightarrow (k + m) \mod n = m \)
Step 1: Solve the congruence
\( (k + m) \mod n = m \Rightarrow k \mod n = 0 \Rightarrow k = 0 \)
Verification: For \( k = 0 \),
\( h(a_0, a_m) = a_{(0 + m) \mod n} = a_m \) which satisfies the condition for all \( m \).
Final Answer: \(\boxed{0}\)
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