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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

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

Answered by MATHEMATICSAM last updated on 01/May/24

$$38(x+y+z)⩽38x+40y+41z$$$$\mathrm{or},38(x+y+z)⩽520$$$$\mathrm{or},x+y+z⩽\frac{520}{38}=13.68...\left(\mathrm{i}\right)$$$$41(x+y+z)⩾38x+40y+41z$$$$\mathrm{or},41(x+y+z)⩾520$$$$\mathrm{or},x+y+z⩾\frac{520}{41}=12.68...\left(\mathrm{ii}\right)$$$$\mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)$$$$12.68⩽x+y+z⩽13.68$$$$x,y,z\in \mathbb{N}$$$$\therefore x+y+z=13\left(\mathrm{Ans}\right)$$

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

$$thanks!$$

Answered by A5T last updated on 01/May/24

$$z=2k\Rightarrow 38x+40y+82k=520$$$$\Rightarrow 19x+20y+41k=260\Rightarrow k⩽6$$$$19x+41k\equiv 0\left(mod10\right)\Rightarrow k\equiv x\left(mod10\right)$$$$-x+k\equiv 0\left(mod20\right)\Rightarrow x=k$$$$\Rightarrow 19k+20y+41k=20y+60k=260$$$$\Rightarrow y+3k=13\Rightarrow y=13-3k$$$$19k+20(13-3k)+41k=260\Rightarrow x=k=0\Rightarrow y=13$$$$\Rightarrow x+y+z=13$$

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