Question

For each \(p > 1\), a sequence \(A_n\) is defined by \(A_0 = 1\) and \(A_n = np + (-1)^n A_{n-1}\) for \(n \geq 1\). For how many integer values of \(p\) is 1000 a term of the sequence?

• A. 8
• B. 7
• C. 5
• D. 4

Hint:

Group the terms by their index mod 4; only two of the four groups can ever equal a value like 1000, and both reduce to divisors of 1001.

Answer 1

The Correct Option is C

Step 1: Compute the first several terms to look for a pattern.
We are given \(A_0 = 1\) and \(A_n = np + (-1)^n A_{n-1}\).
\(A_1 = p + (-1)A_0 = p - 1\)
\(A_2 = 2p + A_1 = 2p + (p - 1) = 3p - 1\)
\(A_3 = 3p - A_2 = 3p - (3p - 1) = 1\)
\(A_4 = 4p + A_3 = 4p + 1\)
\(A_5 = 5p - A_4 = 5p - (4p + 1) = p - 1\)
\(A_6 = 6p + A_5 = 6p + (p - 1) = 7p - 1\)
\(A_7 = 7p - A_6 = 7p - (7p - 1) = 1\)
\(A_8 = 8p + A_7 = 8p + 1\)

Step 2: Group the terms by their remainder when the index is divided by 4.
Looking at the pattern: \(A_1 = p - 1\) and \(A_5 = p - 1\) (indices \(1, 5\), i.e. \(\equiv 1 \bmod 4\)) always equal \(p - 1\).
\(A_3 = 1\) and \(A_7 = 1\) (indices \(\equiv 3 \bmod 4\)) always equal \(1\), which can never be \(1000\).
\(A_2 = 3p - 1\), \(A_6 = 7p - 1\) (indices \(\equiv 2 \bmod 4\)) follow the coefficient pattern \(3, 7, 11, 15, \ldots\) (each 4 more than the last).
\(A_4 = 4p + 1\), \(A_8 = 8p + 1\) (indices \(\equiv 0 \bmod 4\), not counting \(A_0\)) follow the coefficient pattern \(4, 8, 12, \ldots\), and these are always of the "plus 1" form, so they cannot be brought to a "\(cp - 1\)" shape; we focus only on the terms of the form (coefficient) \(\times p - 1\), since 1000 must be reached through one of those.

Step 3: Collect every coefficient c for which some term equals \(cp - 1\).
From Step 2, the achievable coefficients are \(c = 1\) (from \(A_1, A_5, A_9, \ldots\)) and \(c = 3, 7, 11, 15, 19, 23, \ldots\) (from \(A_2, A_6, A_{10}, \ldots\), increasing by 4 each time). So \(c\) is either exactly \(1\), or any number of the form \(4k + 3\) for \(k = 0, 1, 2, \ldots\)

Step 4: Set \(cp - 1 = 1000\), i.e. \(cp = 1001\), and find valid c.
Factorise \(1001 = 7 \times 11 \times 13\). Its full list of divisors is \(1, 7, 11, 13, 77, 91, 143, 1001\).
Check which divisors are allowed values of \(c\) (either \(c = 1\), or \(c \equiv 3 \bmod 4\)):
\(c = 1\): allowed (the special case).
\(c = 7\): \(7 = 4(1) + 3\), allowed.
\(c = 11\): \(11 = 4(2) + 3\), allowed.
\(c = 13\): \(13 = 4(3) + 1\), not allowed.
\(c = 77\): \(77 = 4(19) + 1\), not allowed.
\(c = 91\): \(91 = 4(22) + 3\), allowed.
\(c = 143\): \(143 = 4(35) + 3\), allowed.
\(c = 1001\): \(1001 = 4(250) + 1\), not allowed.
So the allowed divisors are \(c \in \{1, 7, 11, 91, 143\}\), which is 5 values.

Step 5: Turn each allowed c into a value of p.
Since \(cp = 1001\), \(p = \dfrac{1001}{c}\): for \(c = 1\), \(p = 1001\); for \(c = 7\), \(p = 143\); for \(c = 11\), \(p = 91\); for \(c = 91\), \(p = 11\); for \(c = 143\), \(p = 7\). All five values, \(p = 7, 11, 91, 143, 1001\), are integers greater than 1, so all five are valid.

Final Answer:
There are 5 integer values of \(p\) for which 1000 appears somewhere in the sequence; options 8, 7 and 4 do not match the count of divisors of 1001 that satisfy the \(c \equiv 3 \bmod 4\) (or \(c = 1\)) condition.
\[ \boxed{5} \]