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if-28x-30y-31z-360-with-x-y-z-being-positive-integers-find-x-y-z-

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Question Number 226812 by mr W last updated on 15/Dec/25
if 28x+30y+31z=360 with x, y, z being positive integers, find x+y+z=?
$${if}\:\mathrm{28}{x}+\mathrm{30}{y}+\mathrm{31}{z}=\mathrm{360}\:{with}\:{x},\:{y},\:{z} \\ $$$${being}\:{positive}\:{integers},\:{find} \\ $$$${x}+{y}+{z}=? \\ $$
Commented by Ghisom_ last updated on 15/Dec/25
12
$$\mathrm{12} \\ $$
Answered by AgniMath last updated on 16/Dec/25
28x+28y+28z<360 ⇒x+y+z<((360)/(28))≈12.86 31x+31y+31z>360 ⇒x+y+z>((360)/(31))≈11.61 11.61<x+y+z<12.86 x+y+z is integer So x+y+z=12
$$\mathrm{28}{x}+\mathrm{28}{y}+\mathrm{28}{z}<\mathrm{360} \\ $$$$\Rightarrow{x}+{y}+{z}<\frac{\mathrm{360}}{\mathrm{28}}\approx\mathrm{12}.\mathrm{86} \\ $$$$\mathrm{31}{x}+\mathrm{31}{y}+\mathrm{31}{z}>\mathrm{360} \\ $$$$\Rightarrow{x}+{y}+{z}>\frac{\mathrm{360}}{\mathrm{31}}\approx\mathrm{11}.\mathrm{61} \\ $$$$\mathrm{11}.\mathrm{61}<{x}+{y}+{z}<\mathrm{12}.\mathrm{86} \\ $$$${x}+{y}+{z}\:{is}\:{integer} \\ $$$${So}\:{x}+{y}+{z}=\mathrm{12} \\ $$
Commented by mr W last updated on 16/Dec/25
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Commented by A5T last updated on 17/Dec/25
Same method could be used for 28x+30y+32z=359 to get 28x+30y+32z=359<32(x+y+z) and 359>28(x+y+z) to get ((359)/(32))<x+y+z<((359)/(28)) ⇒ x+y+z=12 if x,y,z ∈ Z But no integer solutions actually exist for the equation 28x+30y+32z=359 since 2 divides L.H.S but 2∤359. So, this method does not actually show that a solution exists for positive integers which is what the question is about.
$$\mathrm{Same}\:\mathrm{method}\:\mathrm{could}\:\mathrm{be}\:\mathrm{used}\:\mathrm{for}\: \\ $$$$\mathrm{28x}+\mathrm{30y}+\mathrm{32z}=\mathrm{359}\:\mathrm{to}\:\mathrm{get}\: \\ $$$$\mathrm{28x}+\mathrm{30y}+\mathrm{32z}=\mathrm{359}<\mathrm{32}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right) \\ $$$$\mathrm{and}\:\mathrm{359}>\mathrm{28}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\:\mathrm{to}\:\mathrm{get}\:\frac{\mathrm{359}}{\mathrm{32}}<\mathrm{x}+\mathrm{y}+\mathrm{z}<\frac{\mathrm{359}}{\mathrm{28}} \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{12}\:\mathrm{if}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{Z} \\ $$$$\mathrm{But}\:\mathrm{no}\:\mathrm{integer}\:\mathrm{solutions}\:\mathrm{actually}\:\mathrm{exist}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{28x}+\mathrm{30y}+\mathrm{32z}=\mathrm{359}\:\mathrm{since}\:\mathrm{2}\:\mathrm{divides} \\ $$$$\mathrm{L}.\mathrm{H}.\mathrm{S}\:\mathrm{but}\:\mathrm{2}\nmid\mathrm{359}. \\ $$$$ \\ $$$$\mathrm{So},\:\mathrm{this}\:\mathrm{method}\:\mathrm{does}\:\mathrm{not}\:\mathrm{actually}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{a}\:\mathrm{solution}\:\mathrm{exists}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{what}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{about}. \\ $$
Answered by A5T last updated on 17/Dec/25
Take equation mod 30 ⇒ z−2x≡0(mod 30) ⇒z=30q+2x ; 0<x≤12 because 28×13>360 Similarly, 0<z≤11 ⇒ −24≤z−2x≤11 30q=z−2x ⇒−24≤30q≤11; q∈Z ⇒ q=0 ⇒z=2x ⇒90x+30y=360⇒3x+y=12 ⇒y=12−3x ⇒x+y+z=x+12−3x+2x=12
$$\mathrm{Take}\:\mathrm{equation}\:\mathrm{mod}\:\mathrm{30}\:\Rightarrow\:\mathrm{z}−\mathrm{2x}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{30}\right) \\ $$$$\Rightarrow\mathrm{z}=\mathrm{30q}+\mathrm{2x}\:;\:\mathrm{0}<\mathrm{x}\leqslant\mathrm{12}\:\mathrm{because}\:\mathrm{28}×\mathrm{13}>\mathrm{360} \\ $$$$\mathrm{Similarly},\:\mathrm{0}<\mathrm{z}\leqslant\mathrm{11}\:\Rightarrow\:−\mathrm{24}\leqslant\mathrm{z}−\mathrm{2x}\leqslant\mathrm{11} \\ $$$$\mathrm{30q}=\mathrm{z}−\mathrm{2x}\:\Rightarrow−\mathrm{24}\leqslant\mathrm{30q}\leqslant\mathrm{11};\:\mathrm{q}\in\mathbb{Z}\:\Rightarrow\:\mathrm{q}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{z}=\mathrm{2x}\:\Rightarrow\mathrm{90x}+\mathrm{30y}=\mathrm{360}\Rightarrow\mathrm{3x}+\mathrm{y}=\mathrm{12} \\ $$$$\Rightarrow\mathrm{y}=\mathrm{12}−\mathrm{3x} \\ $$$$\Rightarrow\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{x}+\mathrm{12}−\mathrm{3x}+\mathrm{2x}=\mathrm{12} \\ $$
Commented by A5T last updated on 17/Dec/25
y>0 ⇒ 12−3x>0⇒ x<4 ⇒Integer solutions (x,y,z)=(1,9,2);(2,6,4);(3,3,6)
$$\mathrm{y}>\mathrm{0}\:\Rightarrow\:\mathrm{12}−\mathrm{3x}>\mathrm{0}\Rightarrow\:\mathrm{x}<\mathrm{4}\: \\ $$$$\Rightarrow\mathrm{Integer}\:\mathrm{solutions}\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{1},\mathrm{9},\mathrm{2}\right);\left(\mathrm{2},\mathrm{6},\mathrm{4}\right);\left(\mathrm{3},\mathrm{3},\mathrm{6}\right) \\ $$

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