Sunday Times Teaser 2603 – Pandigital Square

by Des MacHale

Published: 12 August 2012 (link)

I call my telephone number “pandigital”, in the sense that it is a nine-figure number using each of the digits 1 to 9. Amazingly, it is a perfect square. Furthermore, its square root is a five-figure number consisting of five consecutive digits in some order.

It might interest you to know (although you do not need to) that my neighbour’s telephone number is also a nine-figure pandigital perfect square, but his is at least double mine.

With a little logic and not many calculations you should be able to work out my telephone number.

What is it?

One Comment Leave one →

briangladman permalink

Here is my first attempt:


from itertools import permutations

# The first number is less than 5.10^8 so its square root is
# less than 22361.   Since the leading digit is 1 or 2, the
# sequence must be one of 0..4, 1..5 or 2..6. The number is
# divisible by 9 so the square root is divisible by 3 and so
# is the sum of its digits.  The only sequence whose sum is
# a multiple of 3 is 1..5 with the possible square roots of
# 12xxx, 13xxx, 14xxx, 15xxx and 21xxx.

# for each two digit prefix 
for pre in ('12', '13', '14', '15', '21' ):
  # permute the remaining three digits
  for p in permutations(set('12345') - set(pre)):
    # find the number as a string    
    s = str(int(pre + ''.join(p)) ** 2)
    # check it has nine non-zero digits
    if '0' not in s and len(set(s)) == 9:
      print(s)

from itertools import permutations

# The first number is less than 5.10^8 so its square root is

# less than 22361. Since the leading digit is 1 or 2, the

# sequence must be one of 0..4, 1..5 or 2..6. The number is

# divisible by 9 so the square root is divisible by 3 and so

# is the sum of its digits. The only sequence whose sum is

# a multiple of 3 is 1..5 with the possible square roots of

# 12xxx, 13xxx, 14xxx, 15xxx and 21xxx.

# for each two digit prefix

for pre in ('12', '13', '14', '15', '21' ):

# permute the remaining three digits

for p in permutations(set('12345') - set(pre)):

# find the number as a string

s = str(int(pre + ''.join(p)) ** 2)

# check it has nine non-zero digits

if '0' not in s and len(set(s)) == 9:

print(s)

Sunday Times Teaser 2603 – Pandigital Square

by Des MacHale

Published: 12 August 2012 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta