3-Digit Math Problem

Let us solve the following math problem using PyQBPP.

Math Problem: Find all three-digit odd integers whose product of digits is 252.

Let $x$, $y$, and $z$ be the hundreds, tens, and ones digits of the integer, respectively. More specifically:

• $x$ is an integer in $[1, 9]$,
• $y$ is an integer in $[0, 9]$,
• $t$ is an integer in $[0, 4]$,
• $z = 2t + 1$ (so $z$ is odd).

Then the value $v$ of the three-digit integer $xyz$ is

\[\begin{aligned} v&=100x+10y+z \end{aligned}\]

We find all solutions satisfying:

\[\begin{aligned} xyz &= 252 \end{aligned}\]

PyQBPP program

The following PyQBPP program finds all solutions:

import pyqbpp as qbpp

x = qbpp.var("x", integer=(1, 9))
y = qbpp.var("y", integer=(0, 9))
t = qbpp.var("t", integer=(0, 4))
z = 2 * t + 1
v = x * 100 + y * 10 + z

f = qbpp.cons(x * y * z == 252)

f.simplify()
solver = qbpp.ExhaustiveSolver(f)
result = solver.search(best_energy_sols=0)
for sol in result.sols:
    print(sol(v), end=" ")
print()

In this program, x, y, and t are defined as native integer variables with the ranges above. Then z, v, and f are defined as expressions. f declares the condition $xyz = 252$ as a constraint with qbpp.cons(), so its energy takes the minimum value 0 only when the condition holds. We create an Exhaustive Solver instance for f and store all optimal solutions in result.sols.

Because x, y, and t are handled as integers, each solution corresponds to exactly one digit triple (x,y,z), and no triple appears more than once. Printing the value of v for each solution gives:

479 497 667 749 947