How will the pressure of an ideal gas change with an increase in its volume by 2 times

How will the pressure of an ideal gas change with an increase in its volume by 2 times and a decrease in absolute temperature by 3 times?

V2 = 2 * V1.
T2 = T1 / 3.
Р1 / Р2 -?
For an ideal gas, the Mendeleev-Cliperon law is valid: P = n * k * T, where n is the gas concentration, k is the Boltzmann constant, T is the absolute temperature of the gas.
n = N / V, where N is the number of atoms, V is the volume of an ideal gas.
P = N * k * T / V.
Р1 = N * k * T1 / V1.
P2 = N * k * T2 / V2.
Substitute V2 = 2 * V1, T2 = T1 / 3.
P2 = (N * k * T1 / 3) / 2 * V1 = N * k * T1 / 6 * V1 = (1/6) * N * k * T1 / V1 = P1 / 6.
Answer: the pressure of the ideal gas will decrease by 6 times P2 = P1 / 6.