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CAT 2022 Slot 1 — QA Question 12

FactorsEasy

For natural numbers x, y and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is

Answer & solution

Answer: 34

Solution

Easy

Factor each equation. xy+yz=y(x+z)=19xy+yz=y(x+z)=19 is prime, so yy is pinned down at once. Then z(x+1)=51z(x+1)=51 factors a few ways — test each and keep the one giving the smallest xyzxyz.

1

Use the first equation. Since 1919 is prime and x+z≥2x+z\ge 2 for naturals:

xy+yz=y(x+z)=19=1×19⇒ y=1,x+z=19\begin{aligned} &xy+yz=y(x+z)=19=1\times 19\\ &\Rightarrow\ y=1,\quad x+z=19 \end{aligned}
2

Use the second equation with y=1y=1:

yz+xz=z(1+x)=5151=1×51=3×17\begin{aligned} &yz+xz=z(1+x)=51\\ &51=1\times51=3\times17 \end{aligned}
3

Test the factor pairs against x+z=19x+z=19:

z=3, 1+x=17⇒x=16, x+z=19 ✓ ⇒xyz=16⋅1⋅3=48z=17, 1+x=3⇒x=2, x+z=19 ✓ ⇒xyz=2⋅1⋅17=34z=1, 1+x=51⇒x=50, x+z=51 ≠19 ×z=51, 1+x=1⇒x=0 (not natural) ×\begin{aligned} &z=3,\ 1+x=17\Rightarrow x=16,\ x+z=19\ \checkmark\ \Rightarrow xyz=16\cdot1\cdot3=48\\ &z=17,\ 1+x=3\Rightarrow x=2,\ x+z=19\ \checkmark\ \Rightarrow xyz=2\cdot1\cdot17=34\\ &z=1,\ 1+x=51\Rightarrow x=50,\ x+z=51\ \neq19\ \times\\ &z=51,\ 1+x=1\Rightarrow x=0\ \text{(not natural)}\ \times \end{aligned}
4

Pick the minimum. Two valid solutions give 4848 and 3434:

min⁡(48,34)=34\min(48,34)=34
minimum xyz=34\text{minimum }xyz=\mathbf{34}
CAT 2022 Slot 1 QA Q12: For natural numbers x, y and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is — Solution | TheCATExam