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# i may have posted this but I can't find it...

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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
+23140
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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 $11\ \#\ 5$

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

$$\mathbf{11\ \#\ 5 = \ ?}$$

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

Mar 21, 2019
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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