Enigmatic Code
Programming Enigma Puzzles
Enigma 778: Trying triangle
Posted by on 25 March 2024
From New Scientist #1933, 9th July 1994 [link]
Put a digit into each circle and read each side of the triangle clockwise, as a three-figure number. (For example, if I put a 2 in the top left-hand corner and continued, clockwise, 4, 6, 1, 4, 9, then the three-figure numbers would be 246, 614 and 492: notice that 492 is a multiple of 246, but 614 isn’t).
Your job is to choose the digits so that the second and third three-figure numbers are different multiples of the first three figure number.
What are the numbers in your triangle (clockwise from the top left-hand corner)?
[enigma778]
Enigmatic Code 
It is straightforward to use the [[
SubstitutedExpression]] solver from the enigma.py library to solve this puzzle.The following run file executes in 81ms. (Internal runtime of the generated program is 8.2ms).
Run: [ @replit ]
Solution: The triangle looks like this:
And we have:
This is my code:
import itertools for n0, n1, n2, n3, n4, n5 in itertools.product(range(1, 10), repeat=6): p0 = n0 * 100 + n1 * 10 + n2 p1 = n2 * 100 + n3 * 10 + n4 p2 = n4 * 100 + n5 * 10 + n0 if p2 % p0 == 0 and p1 % p0 == 0 and len({p0, p1, p2}) == 3: print(p0, p1, p2)Here’s an extremely performant solution (less then 1 msec on my computer):
import itertools import time t0 = time.perf_counter() for side1 in range(100, 1000): n0 = side1 // 100 n2 = side1 % 10 if n2 > n0: # otherwise, it can't be a multiple of side1 side2 = ((n2 * 100 // side1) + 1) * side1 # try first multiple of side1 starting with n2 if side2 != side1 and side2 < (n2 + 1) * 100: # check whether multiple really starts with n2 n4 = side2 % 10 # n4 is last digit of second side == first digit of third side if n4 > n0: # otherwise, it can't be a multiple of side1 side3 = ((n4 * 100 // side1) + 1) * side1 # try first multiple of p0 starting with n4 if side3 != side1 and side3 < (n4 + 1) * 100: # check whether multiple really starts with n4 n6 = side3 % 10 # last digit of third side if n6 == n0: # check whether this is the same as the first digit of the first side print(side1, side2, side3) print(f"{(time.perf_counter()-t0)*1000} msec")Very fast, one minor error (e.g. for side1 125 the calculated side2 is too high). I adapted your version to this:
from math import ceil # highest multiple * side1 < 1000 for side1 in range(101, 334): n0 = side1 // 100 n2 = side1 % 10 # if n2 is even then n0 must also be even and n2 > n0 # otherwise, it can't be a multiple of side1 if (n2 % 2 or n0 % 2 == 0) and n2 > n0: # try first multiple of side1 starting with n2 side2 = ceil(n2 * 100 / side1) * side1 # check whether multiple really starts with n2 if side2 < (n2 + 1) * 100: # n4 is last digit of second side == first digit of third side n4 = side2 % 10 if n4 > n0: # otherwise, it can't be a multiple of side1 # try first multiple of p0 starting with n4 side3 = ceil(n4 * 100 / side1) * side1 # check whether multiple really starts with n4 if side3 < (n4 + 1) * 100: n6 = side3 % 10 # last digit of third side # check whether this is the same as the first digit of the first side if n6 == n0: # answering the question print("answer:", [n0, (side1 % 100) // 10, n2, (side2 % 100) // 10, n4, (side3 % 100) // 10])Thanks for the bug fix and improvements.
Three times faster according to multiple runs with timeit.
from math import ceil # A - B - C # \ / # F D # \ / # E # E is 0 or 5 if C is equal to 5 (both not allowed) for C in [x for x in range(2, 10) if x != 5]: # if C is even then A and E must also be even for A in range((stp := 1 + (C % 2 == 0)), C, stp): # highest multiple * ABC < 1000 if A > 3 or C * A > 9: break for B in range(10): # break if 2 * (10 * A + B) // 10 > C if 2 * (10 * A + B) > 10 * C + 9: break ABC = 100 * A + 10 * B + C # try first multiple of ABC starting with C if (CDE := ceil(100 * C / ABC) * ABC) // 100 != C: continue E = CDE % 10 # try first multiple of ABC starting with E if E == C or (EFA := ceil(100 * E / ABC) * ABC) // 100 != E: continue if EFA % 10 != A: continue print(f"answer: {A}, {B}, {C}, {(CDE % 100) // 10}, " f"{E} and {(EFA % 100) // 10}")