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Question Number 206924 by mr W last updated on 01/May/24

Commented by mr W last updated on 01/May/24

x, y, z ∈ N

$${x},\:{y},\:{z}\:\in\:\mathbb{N} \\ $$

Answered by MATHEMATICSAM last updated on 01/May/24

38(x + y + z) ≤ 38x + 40y + 41z or, 38(x + y + z) ≤ 520 or, x + y + z ≤ ((520)/(38)) = 13.68 ... (i) 41(x + y + z) ≥ 38x + 40y + 41z or, 41(x + y + z) ≥ 520 or, x + y + z ≥ ((520)/(41)) = 12.68 ... (ii) From (i) and (ii) 12.68 ≤ x + y + z ≤ 13.68 x, y, z ∈ N ∴ x + y + z = 13 (Ans)

$$\mathrm{38}\left({x}\:+\:{y}\:+\:{z}\right)\:\leqslant\:\mathrm{38}{x}\:+\:\mathrm{40}{y}\:+\:\mathrm{41}{z} \\ $$$$\mathrm{or},\:\mathrm{38}\left({x}\:+\:{y}\:+\:{z}\right)\:\leqslant\:\mathrm{520} \\ $$$$\mathrm{or},\:{x}\:+\:{y}\:+\:{z}\:\leqslant\:\frac{\mathrm{520}}{\mathrm{38}}\:=\:\mathrm{13}.\mathrm{68}\:...\:\left(\mathrm{i}\right) \\ $$$$\mathrm{41}\left({x}\:+\:{y}\:+\:{z}\right)\:\geqslant\:\mathrm{38}{x}\:+\:\mathrm{40}{y}\:+\:\mathrm{41}{z} \\ $$$$\mathrm{or},\:\mathrm{41}\left({x}\:+\:{y}\:+\:{z}\right)\:\geqslant\:\mathrm{520} \\ $$$$\mathrm{or},\:{x}\:+\:{y}\:+\:{z}\:\geqslant\:\frac{\mathrm{520}}{\mathrm{41}}\:=\:\mathrm{12}.\mathrm{68}\:...\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{12}.\mathrm{68}\:\leqslant\:{x}\:+\:{y}\:+\:{z}\:\leqslant\:\mathrm{13}.\mathrm{68} \\ $$$${x},\:{y},\:{z}\:\in\:\mathbb{N} \\ $$$$\therefore\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{13}\:\left(\mathrm{Ans}\right) \\ $$

Commented by mr W last updated on 01/May/24

thanks!

$${thanks}! \\ $$

Answered by A5T last updated on 01/May/24

z=2k⇒38x+40y+82k=520 ⇒19x+20y+41k=260⇒k≤6 19x+41k≡0(mod 10)⇒k≡x(mod 10) −x+k≡0(mod 20) ⇒x=k ⇒19k+20y+41k=20y+60k=260 ⇒y+3k=13⇒y=13−3k 19k+20(13−3k)+41k=260⇒x=k=0⇒y=13 ⇒x+y+z=13

$${z}=\mathrm{2}{k}\Rightarrow\mathrm{38}{x}+\mathrm{40}{y}+\mathrm{82}{k}=\mathrm{520} \\ $$$$\Rightarrow\mathrm{19}{x}+\mathrm{20}{y}+\mathrm{41}{k}=\mathrm{260}\Rightarrow{k}\leqslant\mathrm{6} \\ $$$$\mathrm{19}{x}+\mathrm{41}{k}\equiv\mathrm{0}\left({mod}\:\mathrm{10}\right)\Rightarrow{k}\equiv{x}\left({mod}\:\mathrm{10}\right) \\ $$$$−{x}+{k}\equiv\mathrm{0}\left({mod}\:\mathrm{20}\right)\:\:\:\Rightarrow{x}={k} \\ $$$$\Rightarrow\mathrm{19}{k}+\mathrm{20}{y}+\mathrm{41}{k}=\mathrm{20}{y}+\mathrm{60}{k}=\mathrm{260} \\ $$$$\Rightarrow{y}+\mathrm{3}{k}=\mathrm{13}\Rightarrow{y}=\mathrm{13}−\mathrm{3}{k} \\ $$$$\mathrm{19}{k}+\mathrm{20}\left(\mathrm{13}−\mathrm{3}{k}\right)+\mathrm{41}{k}=\mathrm{260}\Rightarrow{x}={k}=\mathrm{0}\Rightarrow{y}=\mathrm{13} \\ $$$$\Rightarrow{x}+{y}+{z}=\mathrm{13} \\ $$

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