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Question Number 10364 by Joel575 last updated on 05/Feb/17

1+2+3+4+5+6+7+8+9+10+11 = x^y   The sum of all possible solutions of x and y is ...

$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10}+\mathrm{11}\:=\:{x}^{{y}} \\ $$$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}\:\mathrm{of}\:{x}\:\mathrm{and}\:{y}\:\mathrm{is}\:... \\ $$

Answered by mrW1 last updated on 05/Feb/17

1+2+3+4+5+6+7+8+9+10+11=((11×12)/2)=66  x^y =66  yln x=ln 66  y=((ln 66)/(ln x))  x∈R,x>0, x≠1  ⇒if x and y can be real numbers,  there are infinite solutions.    if x and y are integers, there is only  one solution:  x=66, y=1  since 66=2×3×11

$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10}+\mathrm{11}=\frac{\mathrm{11}×\mathrm{12}}{\mathrm{2}}=\mathrm{66} \\ $$$${x}^{{y}} =\mathrm{66} \\ $$$${y}\mathrm{ln}\:{x}=\mathrm{ln}\:\mathrm{66} \\ $$$${y}=\frac{\mathrm{ln}\:\mathrm{66}}{\mathrm{ln}\:{x}} \\ $$$${x}\in{R},{x}>\mathrm{0},\:{x}\neq\mathrm{1} \\ $$$$\Rightarrow{if}\:{x}\:{and}\:{y}\:{can}\:{be}\:{real}\:{numbers}, \\ $$$${there}\:{are}\:{infinite}\:{solutions}. \\ $$$$ \\ $$$${if}\:{x}\:{and}\:{y}\:{are}\:{integers},\:{there}\:{is}\:{only} \\ $$$${one}\:{solution}: \\ $$$${x}=\mathrm{66},\:{y}=\mathrm{1} \\ $$$${since}\:\mathrm{66}=\mathrm{2}×\mathrm{3}×\mathrm{11} \\ $$

Commented by Joel575 last updated on 06/Feb/17

thank you sirr

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sirr} \\ $$$$ \\ $$

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