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33-9-25-9-2-19-2-8-




Question Number 125785 by Mathgreat last updated on 13/Dec/20
⊕+⊕+⊕=33  ▲+▲+9▼=25  9▼+▲+2★=19  2⊕+8▼ ∙ ★= ?
$$\oplus+\oplus+\oplus=\mathrm{33} \\ $$$$\blacktriangle+\blacktriangle+\mathrm{9}\blacktrinagledown=\mathrm{25} \\ $$$$\mathrm{9}\blacktrinagledown+\blacktriangle+\mathrm{2}\bigstar=\mathrm{19} \\ $$$$\mathrm{2}\oplus+\mathrm{8}\blacktrinagledown\:\centerdot\:\bigstar=\:? \\ $$$$\:\: \\ $$
Answered by Olaf last updated on 13/Dec/20
 { ((⊕+⊕+⊕ = 33 (1))),((▲+▲+9▼ = 25 (2))),((9▼+▲+2★ = 19 (3))),((2⊕+8▼.★ = ? (4))) :}  (1) : ⊕ = ((33)/3) = 11  (2) : ▲ = ((25−9▼)/2)  (3) : 9▼+((25−9▼)/2)+2★ = 19  ((9▼)/2)+2★ = 19−((25)/2) = ((13)/2)  2★ = ((13−9▼)/2)  (4) : 2⊕+8▼.★ = 22+(4▼)(((13−9▼)/2))  2⊕+8▼.★ = 2(11+13▼−9▼^2 )    If ▼ = 1 ⇒ ★ = 1, ▲ = 4, ⊕ = 11  and 2⊕+8▼.★ = 30
$$\begin{cases}{\oplus+\oplus+\oplus\:=\:\mathrm{33}\:\left(\mathrm{1}\right)}\\{\blacktriangle+\blacktriangle+\mathrm{9}\blacktrinagledown\:=\:\mathrm{25}\:\left(\mathrm{2}\right)}\\{\mathrm{9}\blacktrinagledown+\blacktriangle+\mathrm{2}\bigstar\:=\:\mathrm{19}\:\left(\mathrm{3}\right)}\\{\mathrm{2}\oplus+\mathrm{8}\blacktrinagledown.\bigstar\:=\:?\:\left(\mathrm{4}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)\::\:\oplus\:=\:\frac{\mathrm{33}}{\mathrm{3}}\:=\:\mathrm{11} \\ $$$$\left(\mathrm{2}\right)\::\:\blacktriangle\:=\:\frac{\mathrm{25}−\mathrm{9}\blacktrinagledown}{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\::\:\mathrm{9}\blacktrinagledown+\frac{\mathrm{25}−\mathrm{9}\blacktrinagledown}{\mathrm{2}}+\mathrm{2}\bigstar\:=\:\mathrm{19} \\ $$$$\frac{\mathrm{9}\blacktrinagledown}{\mathrm{2}}+\mathrm{2}\bigstar\:=\:\mathrm{19}−\frac{\mathrm{25}}{\mathrm{2}}\:=\:\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\mathrm{2}\bigstar\:=\:\frac{\mathrm{13}−\mathrm{9}\blacktrinagledown}{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\::\:\mathrm{2}\oplus+\mathrm{8}\blacktrinagledown.\bigstar\:=\:\mathrm{22}+\left(\mathrm{4}\blacktrinagledown\right)\left(\frac{\mathrm{13}−\mathrm{9}\blacktrinagledown}{\mathrm{2}}\right) \\ $$$$\mathrm{2}\oplus+\mathrm{8}\blacktrinagledown.\bigstar\:=\:\mathrm{2}\left(\mathrm{11}+\mathrm{13}\blacktrinagledown−\mathrm{9}\blacktrinagledown^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{If}\:\blacktrinagledown\:=\:\mathrm{1}\:\Rightarrow\:\bigstar\:=\:\mathrm{1},\:\blacktriangle\:=\:\mathrm{4},\:\oplus\:=\:\mathrm{11} \\ $$$$\mathrm{and}\:\mathrm{2}\oplus+\mathrm{8}\blacktrinagledown.\bigstar\:=\:\mathrm{30} \\ $$
Commented by Mathgreat last updated on 13/Dec/20
▲∙▲+9▼=25    ????
$$\blacktriangle\centerdot\blacktriangle+\mathrm{9}\blacktrinagledown=\mathrm{25}\:\:\:\:???? \\ $$

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