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

Try this beautiful problem from Algebra based on positive integers from PRMO 2019. You may use sequential hints to solve the problem.

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

### Key Concepts

Algebra

Integer

sum

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.