Sunday Times Teaser 3296 – Woolly Jumpers

by BRG on November 23, 2025

by Colin Vout

Published Sunday November 23 2025 (link)

Having racked my brains all day trying to devise a suitable Teaser, I woke with my mind racing, just after 4am according to my digital clock. I couldn’t get back to sleep, so I decided to try counting sheep. I imagined them to be numbered starting with the figures shown on the clock, then counting upwards. Each sheep jumped through a particular gap in a fence according to the number of prime factors of its number. Repetitions were counted, so that sheep 405 (=3x3x3x3x5) jumped through the fifth gap.

The last thing I remember noticing before eventually falling asleep was that, for the first time, five consecutive sheep had jumped through the same gap.

What was the number of the last of these five sheep?

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4 Comments Leave one →

BRG permalink

With a neat prime factor count routine from Frits


# count the number of prime factors of <n>
def count_prime_factors(n):
  i, t = 2, 0
  while i * i <= n:
    if n % i:
      i += 1
    else:
      n //= i
      t += 1
  return t + (n > 1)

# the initial sheep number, the number of factors and a repeat count
sheep_nbr, fctr_cnt, rep_cnt = 400, 0, 0
# run until five equal sequential factor counts occur
while rep_cnt < 5:
  sheep_nbr += 1
  # count the factors of each number reached ...
  if (nf := count_prime_factors(sheep_nbr)) != fctr_cnt:
  # ... and restart the repeat count if it has changed
    fctr_cnt, rep_cnt = nf, 0
  rep_cnt += 1
print(f"The last sheep is numbered {sheep_nbr}.")

# count the number of prime factors of <n>

def count_prime_factors(n):

i, t = 2, 0

while i * i <= n:

if n % i:

i += 1

else:

n //= i

t += 1

return t + (n > 1)

# the initial sheep number, the number of factors and a repeat count

sheep_nbr, fctr_cnt, rep_cnt = 400, 0, 0

# run until five equal sequential factor counts occur

while rep_cnt < 5:

sheep_nbr += 1

# count the factors of each number reached ...

if (nf := count_prime_factors(sheep_nbr)) != fctr_cnt:

# ... and restart the repeat count if it has changed

fctr_cnt, rep_cnt = nf, 0

rep_cnt += 1

print(f"The last sheep is numbered {sheep_nbr}.")

JohnZ permalink

Inefficient but interesting (maybe)


from itertools import combinations_with_replacement as cwr

# primes up to 300 (2 x 2 x 300 = 1200)
p = (2, 3, 5, 7, 11, 13, 17)
p = p + tuple(x for x in range(19, 300, 2) if all(x % q for q in p))

# there must be a minimum of 3 primes because within the series of 5
# with 2 primes we have (even, odd, even) or (2 * p1, odd, 2 * p2)
# p2 = p1 + 1 which only occurs if p1 = 2 and p2 = 3
# number of primes from 3 to 8
for np in range(3, 9):
  # use primes to constitute sheep numbers
  sheep = {pxp for ps in cwr(p, np) 
      if 1200 >  (pxp := eval('*'.join(map(str, ps)))) > 406}
  
  # consider starting sheep in series of 5
  for s in range(406, 1196):
    s5 = set(range(s, s + 5))
    if not s5 - sheep:
      print('sheep', s + 4)
      exit()

from itertools import combinations_with_replacement as cwr

# primes up to 300 (2 x 2 x 300 = 1200)

p = (2, 3, 5, 7, 11, 13, 17)

p = p + tuple(x for x in range(19, 300, 2) if all(x % q for q in p))

# there must be a minimum of 3 primes because within the series of 5

# with 2 primes we have (even, odd, even) or (2 * p1, odd, 2 * p2)

# p2 = p1 + 1 which only occurs if p1 = 2 and p2 = 3

# number of primes from 3 to 8

for np in range(3, 9):

# use primes to constitute sheep numbers

sheep = {pxp for ps in cwr(p, np)

if 1200 > (pxp := eval('*'.join(map(str, ps)))) > 406}

# consider starting sheep in series of 5

for s in range(406, 1196):

s5 = set(range(s, s + 5))

if not s5 - sheep:

print('sheep', s + 4)

exit()

John permalink

Python

eval('*'.join(map(str, ps)))

1
2
3

eval('*'.join(map(str, ps)))

is a Heath Robinson-esque way of coding the much faster, more efficient and less interesting

Python

prod(ps)

1
2
3

prod(ps)

after inserting

Python

from math import prod

1
2
3

from math import prod

in the imports section of the code.

Reply

JohnZ permalink

With more information in the printout


from itertools import combinations_with_replacement as cwr

# primes up to 300 (2 x 2 x 300 = 1200)
p = (2, 3, 5, 7, 11, 13, 17)
p = p + tuple(x for x in range(19, 300, 2) if all(x % q for q in p))

# there must be a minimum of 3 primes because within the series of 5
# with 2 primes we have (even, odd, even) or (2 * p1, odd, 2 * p2)
# p2 = p1 + 1 which only occurs if p1 = 1 and p2 = 3
# number of primes from 3 to 8
for np in range(3, 9):
  # use primes to constitute {sheep numbers: factors}
  sheepd = {pxp: ps for ps in cwr(p, np) 
      if 1200 >  (pxp := eval('*'.join(map(str, ps)))) > 406}
  sheep = sheepd.keys()
  # consider starting sheep in series of 5
  for s in range(406, 1196):
    s5 = set(range(s, s + 5))
    if not s5 - sheep:
      for ss in s5:
        print(ss, sheepd[ss])
      print('last sheep:', s + 4)
      exit()

from itertools import combinations_with_replacement as cwr

# primes up to 300 (2 x 2 x 300 = 1200)

p = (2, 3, 5, 7, 11, 13, 17)

p = p + tuple(x for x in range(19, 300, 2) if all(x % q for q in p))

# there must be a minimum of 3 primes because within the series of 5

# with 2 primes we have (even, odd, even) or (2 * p1, odd, 2 * p2)

# p2 = p1 + 1 which only occurs if p1 = 1 and p2 = 3

# number of primes from 3 to 8

for np in range(3, 9):

# use primes to constitute {sheep numbers: factors}

sheepd = {pxp: ps for ps in cwr(p, np)

if 1200 > (pxp := eval('*'.join(map(str, ps)))) > 406}

sheep = sheepd.keys()

# consider starting sheep in series of 5

for s in range(406, 1196):

s5 = set(range(s, s + 5))

if not s5 - sheep:

for ss in s5:

print(ss, sheepd[ss])

print('last sheep:', s + 4)

exit()

Sunday Times Teaser 3296 – Woolly Jumpers

by Colin Vout

Published Sunday November 23 2025 (link)

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