Sunday Times Teaser 2523

by Danny Roth

Published: 30 January 2011 (link)

George and Martha moved to a two-figure house number in Unusual Avenue. The avenue runs from south to north and each house is directly opposite another. No 1 is the southernmost on the west side, then the houses run consecutively up the west side, then southwards back down the east side. So, No 1 is opposite the highest-numbered house. George and Martha’s number is a multiple of the number of the house opposite. The same is true of the five houses their daughters live in, but all their houses have three-figure numbers.

How many houses are there in the avenue?

One Comment Leave one →

Brian Gladman permalink

Here are two different approaches to this one.


from itertools import count

# The number of houses is even, say 2.N, so we want numbers 
# x such that:
# 
#    x = m.(2.N + 1 - x)  ==>  (m + 1).x = m.(2.N + 1)
#
# This makes both x and m even.  We need a number N that
# gives six different values of x and m.


# N is half the number of houses
for N in count(1):
  nbrs = []
  # consider even multiples up to 2.N
  for m in range(2, 2 * N + 1, 2):
    x, r = divmod(m * (2 * N + 1), m + 1)
    if not r:
      # record the solution
      nbrs.extend([x])
      
  # we need at leat six values for x
  if len(nbrs) >= 6:
    
    # compile the pairs of opposite house numbers
    pairs = tuple(tuple(sorted((x, 2 * N + 1 - x))) for x in nbrs)
    
    # we need one pair where the higher number has two digits  
    if any(9 < y < 100 for x, y in pairs):
      
      # and at least five pairs with three digit numbers 
      if len([1 for x, y in pairs[1:] if 99 < max(x, y) < 1000]) >= 5:
        print('{} houses, pairs {}'.format(2 * N, pairs))
        break

from itertools import count

# The number of houses is even, say 2.N, so we want numbers

# x such that:

# x = m.(2.N + 1 - x) ==> (m + 1).x = m.(2.N + 1)

# This makes both x and m even. We need a number N that

# gives six different values of x and m.

# N is half the number of houses

for N in count(1):

nbrs = []

# consider even multiples up to 2.N

for m in range(2, 2 * N + 1, 2):

x, r = divmod(m * (2 * N + 1), m + 1)

if not r:

# record the solution

nbrs.extend([x])

# we need at leat six values for x

if len(nbrs) >= 6:

# compile the pairs of opposite house numbers

pairs = tuple(tuple(sorted((x, 2 * N + 1 - x))) for x in nbrs)

# we need one pair where the higher number has two digits

if any(9 < y < 100 for x, y in pairs):

# and at least five pairs with three digit numbers

if len([1 for x, y in pairs[1:] if 99 < max(x, y) < 1000]) >= 5:

print('{} houses, pairs {}'.format(2 * N, pairs))

break


from itertools import count
from number_theory import divisors

# The number of houses is even, say 2.N, so we want numbers 
# x such that:
# 
#    x = m.(2.N + 1 - x)  ==>  (m + 1).x = m.(2.N + 1)
#
# This makes both x and m even.  We need a number N that
# gives six different values of x and m.

# N is half the number of houses
for N in count(1):
  nbrs = divisors(2 * N + 1)      
  # we need at leat six values for x
  if len(nbrs) >= 6:
    
    # compile the pairs of opposite house numbers
    pairs = tuple(tuple(sorted((x, 2 * N + 1 - x))) for x in nbrs)
    
    # we need one pair where the higher number has two digits  
    if any(9 < y < 100 for x, y in pairs):
      
      # and at least five pairs with three digit numbers 
      if len([1 for x, y in pairs[1:] if 99 < max(x, y) < 1000]) >= 5:
        print('{} houses, pairs {}'.format(2 * N, pairs))
        break