If 2 cdot 4^{2k+1} + 3^{3k+1} = 11t and 2 cdo | vedclass

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(A) Given equations are $2 \cdot 4^{2k+1} + 3^{3k+1} = 11t$ and $2 \cdot 4^{2k+3} + 3^{3k+4} = 11(pt + 3^q)$.We can rewrite the second equation as:$2

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$(16, 3k+1)$

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(A) Given equations are $2 \cdot 4^{2k+1} + 3^{3k+1} = 11t$ and $2 \cdot 4^{2k+3} + 3^{3k+4} = 11(pt + 3^q)$.We can rewrite the second equation as:$2 \cdot 4^{2k+1} \cdot 4^2 + 3^{3k+1} \cdot 3^3 = 11(pt + 3^q)$$16(2 \cdot 4^{2k+1}) + 27(3^{3k+1}) = 11pt + 11 \cdot 3^q$Since $2 \cdot 4^{2k+1} = 11t - 3^{3k+1}$,substitute this into the equation:$16(11t - 3^{3k+1}) + 27(3^{3k+1}) = 11pt + 11 \cdot 3^q$$176t - 16 \cdot 3^{3k+1} + 27 \cdot 3^{3k+1} = 11pt + 11 \cdot 3^q$$176t + 11 \cdot 3^{3k+1} = 11pt + 11 \cdot 3^q$Dividing by $11$:$16t + 3^{3k+1} = pt + 3^q$Comparing the terms,we get $p = 16$ and $q = 3k+1$.Thus,$(p, q) = (16, 3k+1)$.

If $2 \cdot 4^{2k+1} + 3^{3k+1} = 11t$ and $2 \cdot 4^{2k+3} + 3^{3k+4} = 11(pt + 3^q)$,where $k, t \in Z^{+}$,then $(p, q) =$

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