Sunday Times Teaser 2518 – No Title

by Victor Bryant

Published December 26 2010 (link)

For an appropriate Boxing Day exercise, write a digit in each of the boxes (x) so that (counting all the digits from here to the final request) the following statements are true.

_____	number of occurrences of 0	x	_____	total occurrences of 6’s and 7’s	x
	number of occurrences of 1	x		total occurrences of 8’s and 9’s	x
	total occurrences of 2’s and 3’s___	x		a digit you have written in another box___	x
	total occurrences of 4’s and 5’s	x		the average of all your written digits	x
___

What, in order, are the digits in your boxes?

One Comment Leave one →

Brian Gladman permalink


# The total number of digits in the list is the 10 digits 
# plus the 8 yet to be entered - 18 in total. So the sum
# of the answers for the first 6 questions is 18.
#
# There is only 1 '0' digit.
#
# Since the third and higher entries are all two or more, 
# there are only two 1's.
#
# Define sum7 as the sum of the first seven entered digits
# so that the average in box eight is av = (sum7 + av) / 8
# - this shows that sum7 is divisible by seven so 18 plus
# the seventh entry is divisible by seven, which makes the
# two last entries 3

# So we have entries [1, 2, a, b, c, d, 3, 3] with the sum
# of 24, giving a + b + c + d = 15

text = list('''
 Total number of 0's	               1
 Total number of 1's	               2
 Total number of 2's and 3's	       X
 Total number of 4's and 5's	       X
 Total number of 6's and 7's	       X
 Total number of 8's and 9's 	       X
 A digit you put in another box	     3
 The average of the digits entered	 3
''')

# partition an integer (x) into (n) parts 
def partition(x, n, s=[]):
   if n == 1:
      yield s + [x]
   else:
      for i in range(1, x):
         yield from partition(x - i, n - 1, s + [i])

# record position in the text of the unknown values
posn = [i for i, X in enumerate(text) if X == 'X']

# consider possible values for the four unknown entries 
for v in partition(15, 4):
   # enter them into the text
   for p, n in zip(posn, v):
      text[p] = str(n)
   # count the numbers from the text
   n_count = [0] * 4
   for n in range(2, 10):
      n_count[(n // 2) - 1] += text.count(str(n))
   # check if the result is correct
   if v == n_count:
      print(''.join(text))

# The total number of digits in the list is the 10 digits

# plus the 8 yet to be entered - 18 in total. So the sum

# of the answers for the first 6 questions is 18.

# There is only 1 '0' digit.

# Since the third and higher entries are all two or more,

# there are only two 1's.

# Define sum7 as the sum of the first seven entered digits

# so that the average in box eight is av = (sum7 + av) / 8

# - this shows that sum7 is divisible by seven so 18 plus

# the seventh entry is divisible by seven, which makes the

# two last entries 3

# So we have entries [1, 2, a, b, c, d, 3, 3] with the sum

# of 24, giving a + b + c + d = 15

text = list('''

Total number of 0's 1

Total number of 1's 2

Total number of 2's and 3's X

Total number of 4's and 5's X

Total number of 6's and 7's X

Total number of 8's and 9's X

A digit you put in another box 3

The average of the digits entered 3

''')

# partition an integer (x) into (n) parts

def partition(x, n, s=[]):

if n == 1:

yield s + [x]

else:

for i in range(1, x):

yield from partition(x - i, n - 1, s + [i])

# record position in the text of the unknown values

posn = [i for i, X in enumerate(text) if X == 'X']

# consider possible values for the four unknown entries

for v in partition(15, 4):

# enter them into the text

for p, n in zip(posn, v):

text[p] = str(n)

# count the numbers from the text

n_count = [0] * 4

for n in range(2, 10):

n_count[(n // 2) - 1] += text.count(str(n))

# check if the result is correct

if v == n_count:

print(''.join(text))

Sunday Times Teaser 2518 – No Title

by Victor Bryant

Published December 26 2010 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta