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Bunuel
Consider a sequence \(a_1\), \(a_2\), \(a_3\), \(a_4\), ..., \(a_{n-2}\), \(a_{n-1}\) and \(a_n\) defined as \(a_n\) is the remainder when \(a_{n-2}\) is divided by \(a_{n-1}\) for all \(n\geq{3}\). If \(a_4\) is the first zero term, \(a_3\) is 6 then what is value of \(a_1\)?

A. 48
B. 49
C. 50
D. 51
E. 52
\( a_4 = 0 \)
\( a_3 = 6 \)

\( a_2 = a_3*k + a_4 \)
\( a_2 = 6k \)

\( a_1 = a_2*p + a_3 \)
\( a_1 = 6kp + 6 \)
\( a_1 = 6(kp + 1)\)

Since \(a_1\) should be multiple of 6, only option possible from the given choices is 48.

Answer: A
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