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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$(16, 3k+1)$

Vedclass · Answer

(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 $(p, q) =$

Similar Questions

For all integers $n \geq 1$,which of the following is divisible by $9$?

For any integer $n \geq 1$,the remainder when the expression $n^5-5n^3+4n+139$ is divided by $120$ is

If $a, b$ and $n$ are natural numbers,then $a^{2n-1} + b^{2n-1}$ is divisible by

If $[x]$ denotes the greatest integer function of $x$,then the number of positive integral divisors of $\left[(2+\sqrt{3})^5\right]$ is

The remainder when $(11)^{1011} + (1011)^{11}$ is divided by $9$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module