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For all real numbers $r$ and $s$, define the mathematical operation $\#$ such that the following conditions apply: $r\ \#\ 0 = r, r\ \#\ s = s\ \#\ r$, and $(r + 1)\ \#\ s = (r\ \#\ s) + s + 1$. What is the value of $11\ \#\ 5$?
 

For all real numbers \(r\) and \(s\), define the mathematical operation \(\#\) such that the following conditions apply: \(r\ \#\ 0 = r, r\ \#\ s = s\ \#\ r\), and \((r + 1)\ \#\ s = (r\ \#\ s) + s + 1\). What is the value of \(11\ \#\ 5\)?

 Mar 21, 2019
 #1
avatar+22896 
+3

For all real numbers $r$ and $s$, define the mathematical operation $\#$ such that the following conditions apply:

$r\ \#\ 0 = r, r\ \#\ s = s\ \#\ r$, and $(r + 1)\ \#\ s = (r\ \#\ s) + s + 1$.

What is the value of $11\ \#\ 5$

 

 

\(\begin{array}{|rcll|} \hline \mathbf{(r + 1)\ \#\ s }& \mathbf{=} & \mathbf{(r\ \#\ s) + s + 1} \quad \text{ or } \quad \mathbf{r\ \#\ s = \Big((r-1)\ \#\ s\Big) + s + 1} \\ \hline \\ r\ \#\ s &=& \Big((r-1)\ \#\ s\Big) + s + 1 \quad | \quad (r-1)\ \#\ s = \Big((r-2)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-2)\ \#\ s\Big) + s + 1 + s + 1 \\ &=& \Big((r-2)\ \#\ s\Big) + 2s + 2 \quad | \quad (r-2)\ \#\ s = \Big((r-3)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-3)\ \#\ s\Big) + s + 1 + 2s + 2 \\ &=& \Big((r-3)\ \#\ s\Big) + 3s + 3 \quad | \quad (r-3)\ \#\ s = \Big((r-4)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-4)\ \#\ s\Big) + s + 1 + 3s + 3 \\ &=& \Big((r-4)\ \#\ s\Big) + 4s + 4 \\ \ldots \\ \mathbf{r\ \#\ s} &\mathbf{=}& \mathbf{ \Big((r-n)\ \#\ s\Big) + ns + n} \qquad n\in \mathbb{Z} \\ \hline \end{array} \)

 

\(\mathbf{11\ \#\ 5 = \ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{r\ \#\ s} &\mathbf{=}& \mathbf{ \Big((r-n)\ \#\ s\Big) + ns + n} \quad &| \quad r=n=11,\ s=5 \\ 11\ \#\ 5 & = & \Big((11-11)\ \#\ 5\Big) + 11\cdot 5 + 11 \\ 11\ \#\ 5 & = & \Big(0\ \#\ 5\Big) + 11\cdot 5 + 1 \quad &| \quad 0\ \#\ 5 = 5\ \#\ 0 \\ 11\ \#\ 5 & = & \Big(5\ \#\ 0\Big) + 11\cdot 5 + 11 \quad &| \quad 5\ \#\ 0 = 5 \\ 11\ \#\ 5 & = & 5 + 11\cdot 5 + 11 \\ \mathbf{11\ \#\ 5} & \mathbf{=} & \mathbf{71} \\ \hline \end{array}\)

 

laugh

 Mar 21, 2019
 #2
avatar+102791 
+1

Thanks Heureka.  :)

 

Guest if you join all of  your posts will be very easy to keep track of. 

So you should join.

 Mar 21, 2019

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