Sunday Times Teaser 2667 – Prime Time

by Peter G Chamberlain

I wrote an A next to one number on a diagram of a clock face and then moved a number of hour positions clockwise and wrote a B.

I continued in this way using the same number of clockwise steps to write all the following letters so that they each had a corresponding number.

I noticed that the two numbers with three letters by them were both prime.

I also noticed that if I wrote the numbers for PRIME in order to form a long number, I obtained a six-digit prime.

What numbers correspond to A and B respectively?

3 Comments Leave one →

brian gladman permalink

This uses my prime class provided under the ‘Python’ menu above.


from primes import Primes

# The number on which A is placed must be prime since 
# A, M and Y are on the same number whatever the step
for pos_a in (2, 3, 5, 7, 11):
  # the number of steps clockwise between consecutive letters
  # - for only 2 numbers to have three letters means that the
  # step is co-prime with 12
  for step in (1, 5, 7, 11):
    
    # B, N and Z are on the same number so it must be prime
    if (pos_a + step) % 12 in (2, 3, 5, 7, 11):
      
      # l2n maps letters to numbers for the 26 letters
      l2n = {chr(ord('A') + i):(pos_a + i * step) % 12 for i in range(26)}
        
      # find the value of 'PRIME' (noting that 0 on clock is labelled 12)
      nbr = ''.join(str(l2n[c] if l2n[c] else 12) for c in 'PRIME')
      # check that this is a six digit prime
      if len(nbr) == 6 and int(nbr) in Primes():
        print('A and B correspond to {} and {} respectively (PRIME = {}).'
              .format(l2n['A'], l2n['B'], nbr))

from primes import Primes

# The number on which A is placed must be prime since

# A, M and Y are on the same number whatever the step

for pos_a in (2, 3, 5, 7, 11):

# the number of steps clockwise between consecutive letters

# - for only 2 numbers to have three letters means that the

# step is co-prime with 12

for step in (1, 5, 7, 11):

# B, N and Z are on the same number so it must be prime

if (pos_a + step) % 12 in (2, 3, 5, 7, 11):

# l2n maps letters to numbers for the 26 letters

l2n = {chr(ord('A') + i):(pos_a + i * step) % 12 for i in range(26)}

# find the value of 'PRIME' (noting that 0 on clock is labelled 12)

nbr = ''.join(str(l2n[c] if l2n[c] else 12) for c in 'PRIME')

# check that this is a six digit prime

if len(nbr) == 6 and int(nbr) in Primes():

print('A and B correspond to {} and {} respectively (PRIME = {}).'

.format(l2n['A'], l2n['B'], nbr))

Arthur Vause permalink

Here is my version, using a simple sieve to generate primes:


plist = [False,False,True]+[True,False]*500000
for n in xrange(3,1000,2):
  if plist[n]: plist[n*n::2*n]= [False]*len(plist[n*n::2*n])
primes=set(x for x in xrange(1000000) if plist[x])

for h in (1,5,7,11):
  for A in xrange(1,13):
    B,P,R,I,M,E,Y,Z = ( str(((A + (i-1)*h)-1)%12+1)
                          for i in (2,16,18,9,13,5,25,26))
    if int(Y) in primes and int(Z) in primes:
      if int(P+R+I+M+E) in primes: 
         print "increment=",h, "P,R,I,M,E=",P,R,I,M,E, "PRIME=",P+R+I+M+E
         print "A,B=",A,B

plist = [False,False,True]+[True,False]*500000

for n in xrange(3,1000,2):

if plist[n]: plist[n*n::2*n]= [False]*len(plist[n*n::2*n])

primes=set(x for x in xrange(1000000) if plist[x])

for h in (1,5,7,11):

for A in xrange(1,13):

B,P,R,I,M,E,Y,Z = ( str(((A + (i-1)*h)-1)%12+1)

for i in (2,16,18,9,13,5,25,26))

if int(Y) in primes and int(Z) in primes:

if int(P+R+I+M+E) in primes:

print "increment=",h, "P,R,I,M,E=",P,R,I,M,E, "PRIME=",P+R+I+M+E

print "A,B=",A,B

ahmet cetinbudaklar permalink

E must be either 1 or 3 or 7 or 9 or 11 to let PR?ME be a prime number as well as step lengths must not necessarily be factors of 12 , then it is a matter of trial and error to get the solution.

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Sunday Times Teaser 2667 – Prime Time

by Peter G Chamberlain

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