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# Problem on Positive Integers | PRMO-2019 | Problem 26

Try this beautiful problem from Algebra PRMO 2019 based on Positive Integers.

## Positive Integers | PRMO | Problem 26

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

• 24
• 50
• 29
• 34

Algebra

Integer

sum

## Check the Answer

PRMO-2019, Problem 26

Higher Algebra by Hall and Knight

## Try with Hints

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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