Sunday Times Teaser 3168 – Number Cruncher

by BRG on June 9, 2023

by Colin Vout

Published Sunday June 11 2023 (link)

The gaslight glittered on the polished brass of the machine. “My Number Cruncher can calculate mechanically the result of a calculation on three positive whole numbers used once each, provided it is restricted to plus, times and brackets operations,” declared the Engineer. He opened the delicate cover and made meticulous adjustments. “There, I have disposed it for one particular expression. Please put it to the test.”

I chose three numbers, all greater than 1, and rotated the three dials to those positions. Then I cranked the handle until a delicate bell rang, and the result was indicated as 451. I made one more try, using the same three numbers although in a different order; this time the machine yielded 331.

In ascending order, what were the three numbers I selected?

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BRG permalink


# Neither a + b + c nor a x b x c change when the numbers are 
# rearranged so brackets are needed around two of the numbers
#
# The expression (a + b) x c = 331 is not possible because 
# 331 is a prime and all of the numbers are greater than 1
#
# So the expression is (a x b) + c = P and (a x c) + b = Q
# with P = 331 (a prime) and Q = 451 (which is not prime)
#
# Some further algebra shows that
#
#    (a + 1).(c + b) = Q + P
#    (a - 1).(c - b) = Q - P
#    a.c + b = Q  ==>  [Q + (c - b)] / (a + 1)

P, Q = 331, 451
D, S = Q - P, Q + P

for am1 in range(1, D + 1):
  cmb, r = divmod(D, am1)
  if not r:
    a = am1 + 1
    c = (Q + cmb) // (a + 1)
    b = c - cmb
    if (a + 1) * (b + c) == S:
      print(f'{a} {b} {c}')

# Neither a + b + c nor a x b x c change when the numbers are

# rearranged so brackets are needed around two of the numbers

# The expression (a + b) x c = 331 is not possible because

# 331 is a prime and all of the numbers are greater than 1

# So the expression is (a x b) + c = P and (a x c) + b = Q

# with P = 331 (a prime) and Q = 451 (which is not prime)

# Some further algebra shows that

# (a + 1).(c + b) = Q + P

# (a - 1).(c - b) = Q - P

# a.c + b = Q ==> [Q + (c - b)] / (a + 1)

P, Q = 331, 451

D, S = Q - P, Q + P

for am1 in range(1, D + 1):

cmb, r = divmod(D, am1)

if not r:

a = am1 + 1

c = (Q + cmb) // (a + 1)

b = c - cmb

if (a + 1) * (b + c) == S:

print(f'{a} {b} {c}')

John Z permalink

In your comment line 8, how do you exclude (b x c) + a = Q as a possibility without after-the-fact knowledge?

Reply
- BRG permalink
  
  Hi John, If the solution for swapping b with c is (a, b, c) = (x, y, z) then
  choosing instead to swap a with c will give the solution (a, b, c) = (y, x, z).
  So we can choose either swap since the results will be the same. I used
  the definition that produced a < b < c the avoid the need for sorting.
  
  Reply

JohnZ permalink


'''
Target values are 451 and 331. Equations of the form:
a * b * c = 451 or 331
a + b + c
are excluded because they give a unique result regardless of the order of the
operands. Equations of the form:
(a + b) * c
a * (b + c) are excluded because 331 is prime.
This leaves:
(a * b) + c
'''

#consider values for <a> and <b> such that a * b < 330
for a in range(2, 165):
  for b in range(a + 1, 166):
    if a * b > 330:
      break
    c = 331 - a * b
    if (a * c) + b == 451 or (b * c) + a == 451:
      print(a, b, c)

'''

Target values are 451 and 331. Equations of the form:

a * b * c = 451 or 331

a + b + c

are excluded because they give a unique result regardless of the order of the

operands. Equations of the form:

(a + b) * c

a * (b + c) are excluded because 331 is prime.

This leaves:

(a * b) + c

'''

#consider values for <a> and <b> such that a * b < 330

for a in range(2, 165):

for b in range(a + 1, 166):

if a * b > 330:

break

c = 331 - a * b

if (a * c) + b == 451 or (b * c) + a == 451:

print(a, b, c)

Sunday Times Teaser 3168 – Number Cruncher

by Colin Vout

Published Sunday June 11 2023 (link)

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