Problem on Positive Integers | PRMO-2019 | Problem 26

Positive integers x,y,z satisfy xy+z=160 compute smallest possible value of x+yz.

x+yz=\(\frac{160-z}{y}\)+yz

=\(\frac{160}{y}+\frac{z(y^{2}-1)}{y}=\frac{160-z}{y}+\frac{zy^{2}}{y}=\frac{160-z}{y}+zy\)

for particular value of z, \(x+yz \geq 2\sqrt{z(160-z)}\)

or, least value=\(2\sqrt{z(160-z)}\) but an integer also

for least value z is also

case I z=1, \(x+yz=\frac{159}{y}+y\) or, min value at y=3 which is 56

case II z=2, \(x+yz=\frac{158}{y}+2y\) or, min value at y =2 which is 83 (not taken)

case III z=3, \(x+yz=\frac{157}{y}+3y\) or, min value at y=1 which is 160 (not taken)

case IV z=4, \(x+yz=\frac{156}{y}+4y\) or, min at y=6 which is 50 (taken)

case V z=5, \(x+yz=\frac{155}{y}+5y\) or, minimum value at y=5 which is 56 (not taken)

case VI z=6, \(x+yz=\frac{154}{y}+6y\) \( \geq 2\sqrt{924}\)>50

smallest possible value =50.

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