Sunday Times Teaser 2676 – New Diary

by Grahamn Smithers

In a 3 by 2 rectangular grid there are 12 junctions at which lines either meet or cross. When defined as any two junctions that are joined by straight lines in the grid, we can find 30 ‘junction pairs’.

For many years, diaries have been published using various size grids and the 1998 diary was special because its grid had exactly 1998 such junction pairs.

For one year in the next decade (starting in 2014), a diary can be published for which the number of such junction pairs is equal to the year of publication.

(a) What was the grid size of the 1998 diary?

(b) what is the year of the next special diary?

2 Comments Leave one →

brian gladman permalink


# Let the grid be m (horizontal) by n (vertical) with m + 1 vertical
# lines and n + 1 horizontal lines. Each horizontal line contributes
# 1 + 2 + ... + m junction pairs so there are  (n + 1).m.(m + 1) / 2
# in total. The same is true for the vertical lines but with m and n
# exchanged so the overall total is (m + 1).(n + 1).(m + n) / 2.

# the maximum grid dimensions considered
g_max = 40

sol = set()
# horizontal grid size
for m in range(1, g_max):
  # vertical grid size
  for n in range(m, g_max):
    # calculate the number of pairs
    p = (m + 1) * (n + 1) * (m + n) // 2
    # for question (a)
    if p == 1998:
      print('(a) {} by {}'.format(m, n))
    # for question (b)
    if 2013 < p < 2025:
      sol.add(p)
# output solution(s) for question (b)
for y in sol:
  print('(b) {}'.format(y))

# Let the grid be m (horizontal) by n (vertical) with m + 1 vertical

# lines and n + 1 horizontal lines. Each horizontal line contributes

# 1 + 2 + ... + m junction pairs so there are (n + 1).m.(m + 1) / 2

# in total. The same is true for the vertical lines but with m and n

# exchanged so the overall total is (m + 1).(n + 1).(m + n) / 2.

# the maximum grid dimensions considered

g_max = 40

sol = set()

# horizontal grid size

for m in range(1, g_max):

# vertical grid size

for n in range(m, g_max):

# calculate the number of pairs

p = (m + 1) * (n + 1) * (m + n) // 2

# for question (a)

if p == 1998:

print('(a) {} by {}'.format(m, n))

# for question (b)

if 2013 < p < 2025:

sol.add(p)

# output solution(s) for question (b)

for y in sol:

print('(b) {}'.format(y))

ahmet cetinbudaklar permalink

If the number of junctions=(n+m)(n+1)(m+1)/2

Hence for 1998 1998×2=3996=37x36x3=(35+2)(35+1)(2+1) and
for 2016 2016×2=4032=28x24x6=(23+5)(23+1)(5+1)
for 2025 2025×2=4050=25x18x9=(17+8)(17+1)(8+1)
and may be more examples to follw…

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Sunday Times Teaser 2676 – New Diary

by Grahamn Smithers

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