N is a 5-digit duodecimal prime number which consists of precisely two distinct
nonzero base-12 digits.
Determine the smallest and largest value of N.
*** N cannot admit of any leading zeros.
If the problem is interpreted as precisely two distinct nonzero base-12 digits and only nonzero digits are allowed, then:
There are 171 such 5-digit duodecimal primes
Smallest:
11151 which is 22669 in base 10
Largest: BBBB7 which is 248827 in base 10
In what is probably a bizarre coincidence, if you eliminate the constraint of the digits all being nonzero, then instead of 171 solutions, you get 172; and the sole solution containing zeros is 10011 which is then the smallest.
If the problem is interpreted as precisely two distinct nonzero base-12 digits but in addition to those nonzero digits, some zero digits are allowed, (
although I do not believe this is the intended interpretation) then:
There are 554 such 5-digit duodecimal primes
Smallest: 10007 which is 20743 in base 10
Largest: BBBB7 which is 248827 in base 10
----------
def isprime(n):
'''check if integer n is a prime'''
n = abs(int(n))
if n < 2:
return False
if n == 2:
return True
if not n & 1:
return False
for x in range(3, int(n**0.5)+1, 2):
if n % x == 0:
return False
return True
def base2base(n,a,b):
""" input n which is a string of the number to be changed
'a' is an integer representing the current base
to be changed to base b
"""
def dec2base(i,base):
""" INPUT integer in base 10, return string
of Base base equivalent. """
convertString = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'
if i < base:
return convertString[i]
else:
return dec2base(i//base,base) + convertString[i%base]
if a == 10:
return dec2base(int(n),b)
if b == 10:
return int(str(n),a)
elif b != 10:
return base2base(int(str(n),a),10,b)
lo = base2base('10000',12,10)
hi = base2base('BBBBB',12,10)
zerosAllowed = True
count = 0
count0 = 0
ans = []
answithzeros = []
for zerosAllowed in [True,False]:
for n in range(lo,hi):
if not isprime(n):
continue
n12 = base2base(n,10,12)
if not zerosAllowed:
if '0' in n12:
continue
if len(set(n12)) != 2:
continue
ans.append(n12)
count += 1
print(n12)
else:
if '0' in n12:
if len(set(n12)) != 3:
continue
elif '0' not in n12:
if len(set(n12)) != 2:
continue
answithzeros.append(n12)
count0 += 1
print(n12)
print('no zeros',count)
print('zeros allowed',count0)
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Posted by Larry
on 2023-03-24 09:25:45 |
Computer solution
