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Question Number 147231 by mathdanisur last updated on 19/Jul/21

if 3^(a+2) = 5^(3b−1) = 7^(3−2c) find a∙b∙c = ?

$${if}\:\:\:\mathrm{3}^{\boldsymbol{{a}}+\mathrm{2}} \:=\:\mathrm{5}^{\mathrm{3}\boldsymbol{{b}}−\mathrm{1}} \:=\:\mathrm{7}^{\mathrm{3}−\mathrm{2}\boldsymbol{{c}}} \\ $$$${find}\:\:\:\boldsymbol{{a}}\centerdot\boldsymbol{{b}}\centerdot\boldsymbol{{c}}\:=\:? \\ $$

Answered by Olaf_Thorendsen last updated on 19/Jul/21

3^(a+2) = 5^(3b−1) = 7^(3−2c) The last digit of 5^(3b−1) is 1 if 3b−1 = 0 or 5 if 3b−1 ≠ 0 But the last digit of 3^(a+2) and 7^(3−2c) canno′t be 5. The only solution is 3b−1 = 0 then 3^(a+2) = 5^(3b−1) = 7^(3−2c) = 1 and a = −2, b = (1/3), c = (3/2) abc = −2×(1/3)×(3/2) = −1

$$\mathrm{3}^{{a}+\mathrm{2}} \:=\:\mathrm{5}^{\mathrm{3}{b}−\mathrm{1}} \:=\:\mathrm{7}^{\mathrm{3}−\mathrm{2}{c}} \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{5}^{\mathrm{3}{b}−\mathrm{1}} \:\mathrm{is}\:\mathrm{1}\:\mathrm{if}\:\mathrm{3}{b}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{or}\:\mathrm{5}\:\mathrm{if}\:\mathrm{3}{b}−\mathrm{1}\:\neq\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{3}^{{a}+\mathrm{2}} \:\:\mathrm{and}\:\mathrm{7}^{\mathrm{3}−\mathrm{2}{c}} \\ $$$$\mathrm{canno}'\mathrm{t}\:\mathrm{be}\:\mathrm{5}. \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{3}{b}−\mathrm{1}\:=\:\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{3}^{{a}+\mathrm{2}} \:=\:\mathrm{5}^{\mathrm{3}{b}−\mathrm{1}} \:=\:\mathrm{7}^{\mathrm{3}−\mathrm{2}{c}} \:=\:\mathrm{1}\:\mathrm{and} \\ $$$${a}\:=\:−\mathrm{2},\:{b}\:=\:\frac{\mathrm{1}}{\mathrm{3}},\:{c}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$${abc}\:=\:−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{2}}\:=\:−\mathrm{1} \\ $$

Commented by mathdanisur last updated on 19/Jul/21

cool Ser thank you

$${cool}\:{Ser}\:{thank}\:{you} \\ $$

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