Sunday Times Teaser 2642 – New World Order

by H Bradley and C Higgins

A text book with pages 1 to 500 inclusive has three sections, Arithmetic, Algebra and Geometry, each starting on a new page.

The sum of the page numbers in the Algebra section is double the same sum for the Geometry section and the latter is over twice as long as the Arithmetic section.

How many pages are there in the Arithmetic section?

2 Comments Leave one →

Brian Gladman permalink

Here is a straightforward solution:


# the number of pages in the arithmetic section
for ar_pages in range(1, 167):
  # the sum of its page numbers
  ar_sum = ar_pages * (ar_pages + 1) // 2

  # the number of pages in the geometry section
  for ge_pages in range(2 * ar_pages + 1, 334):
    # the last page number in the algebra section
    al_end = 500 - ge_pages

    # the sums of the page numbers in the algebra
    # and geometry sections
    t = al_end * (al_end + 1) // 2
    al_sum, ge_sum = t - ar_sum, 250 * 501 - t

    # the algebra page sum is twice the geometry page sum
    if al_sum == 2 * ge_sum:
      print('The arithmetic section has {:d} pages.'.format(ar_pages))

# the number of pages in the arithmetic section

for ar_pages in range(1, 167):

# the sum of its page numbers

ar_sum = ar_pages * (ar_pages + 1) // 2

# the number of pages in the geometry section

for ge_pages in range(2 * ar_pages + 1, 334):

# the last page number in the algebra section

al_end = 500 - ge_pages

# the sums of the page numbers in the algebra

# and geometry sections

t = al_end * (al_end + 1) // 2

al_sum, ge_sum = t - ar_sum, 250 * 501 - t

# the algebra page sum is twice the geometry page sum

if al_sum == 2 * ge_sum:

print('The arithmetic section has {:d} pages.'.format(ar_pages))

and here is another solution based on further analysis leading to a Pell equation (see here):


# Let there be r pages in the arithmetic section and let the end page of
# the algebra section be e. The three sums are then:
#
# (1) arithmetic section sum = r * (r + 1) // 2
# (2) algebra section sum    = e * (e + 1) // 2 - r * (r + 1) // 2
# (3) geometry section sum   =   500 * 501 // 2 - e * (e + 1) // 2
#
# Since (2) is twice (3):
#
#    3 * e * (e + 1) - r * (r + 1) == 1000 * 501
#
# which can be put in the form:
#
#    (2 * r + 1) ** 2 - 3 * (2 * e + 1) ** 2 == -2 * (7 * 11 * 13) ** 2
#
# Also 500 - e > 2r ==> (e = r + x) ==> r < (500 - x) / 3 ==> r < 167

for r in range(167):
  div, rem = divmod((2 * r + 1) ** 2 + 2 * (7 * 11 * 13) ** 2, 3)
  if not rem and round(div ** 0.5) ** 2 == div:
    print('The arithmetic section has {:d} pages.'.format(r))

# Let there be r pages in the arithmetic section and let the end page of

# the algebra section be e. The three sums are then:

# (1) arithmetic section sum = r * (r + 1) // 2

# (2) algebra section sum = e * (e + 1) // 2 - r * (r + 1) // 2

# (3) geometry section sum = 500 * 501 // 2 - e * (e + 1) // 2

# Since (2) is twice (3):

# 3 * e * (e + 1) - r * (r + 1) == 1000 * 501

# which can be put in the form:

# (2 * r + 1) ** 2 - 3 * (2 * e + 1) ** 2 == -2 * (7 * 11 * 13) ** 2

# Also 500 - e > 2r ==> (e = r + x) ==> r < (500 - x) / 3 ==> r < 167

for r in range(167):

div, rem = divmod((2 * r + 1) ** 2 + 2 * (7 * 11 * 13) ** 2, 3)

if not rem and round(div ** 0.5) ** 2 == div:

print('The arithmetic section has {:d} pages.'.format(r))

Frits permalink

Nice analysis. The following program covers the full solution space.


from math import ceil

# Let there be r pages in the arithmetic section and let the end page of
# the algebra section be e. The three sums are then:
#
# (1) arithmetic section sum = r * (r + 1) // 2
# (2) algebra section sum    = e * (e + 1) // 2 - r * (r + 1) // 2
# (3) geometry section sum   = 500 * 501 // 2 - e * (e + 1) // 2
#
# Since (2) is twice (3):
#
#    3 * e * (e + 1) - r * (r + 1) == 1000 * 501
#
# which can be put in the form:
#
#    (2 * r + 1) ** 2 - 3 * (2 * e + 1) ** 2 == -2 * (7 * 11 * 13) ** 2
#
# so ((2 * r + 1) ** 2 + 2 * (7 * 11 * 13) ** 2) / 3 must be a square 
 

N = 2 * (7 * 11 * 13) ** 2

# ((2 * r + 1)^2 + N) / 3 is n^2 so 3 * n^2 - N = (2 * r + 1)^2
n = ceil((N / 3)**.5)   
while True:
  r, rem = divmod((3 * n**2 - N) ** 0.5 - 1, 2)
  # int(499 / 3) is maximum for r
  if r > 166: break
  if not rem: 
    print("answer:", int(r))
    
  n += 1

from math import ceil

# Let there be r pages in the arithmetic section and let the end page of

# the algebra section be e. The three sums are then:

# (1) arithmetic section sum = r * (r + 1) // 2

# (2) algebra section sum = e * (e + 1) // 2 - r * (r + 1) // 2

# (3) geometry section sum = 500 * 501 // 2 - e * (e + 1) // 2

# Since (2) is twice (3):

# 3 * e * (e + 1) - r * (r + 1) == 1000 * 501

# which can be put in the form:

# (2 * r + 1) ** 2 - 3 * (2 * e + 1) ** 2 == -2 * (7 * 11 * 13) ** 2

# so ((2 * r + 1) ** 2 + 2 * (7 * 11 * 13) ** 2) / 3 must be a square

N = 2 * (7 * 11 * 13) ** 2

# ((2 * r + 1)^2 + N) / 3 is n^2 so 3 * n^2 - N = (2 * r + 1)^2

n = ceil((N / 3)**.5)

while True:

r, rem = divmod((3 * n**2 - N) ** 0.5 - 1, 2)

# int(499 / 3) is maximum for r

if r > 166: break

if not rem:

print("answer:", int(r))

n += 1

Sunday Times Teaser 2642 – New World Order

by H Bradley and C Higgins

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