An eight-digit number is said to be 'robust' if it meets both of the following conditions:
(i) None of its digits is 0.
(ii) The difference between two consecutive digits is 4 or 5.
Answer the following questions:
(a) How many robust numbers are there?
(b) A robust number is said to be 'super robust' if all of its digits are distinct. Calculate the sum of all the super robust numbers.
The first number can be any of 9 choices. There are always 2 choices for subsequent digits. 9 * 2^7
To form a super robust number start with any of the 9 digits, then either go up 5 or down 4 (there is always only one way to follow this instruction); or go up 4 or down 5. So for each starting digit, there are 2 super robust numbers for a total of 18. The average digit is 5, so the average of all super robust numbers is 55555555.
(i) There are 9*2^7 = 1152 Robust numbers
(ii) There are 18 super robust numbers
and their sum is 18 * 55555555 = 999999990
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Program output to confirm:
1152 18
999999990
[15948372, 16273849, 26159483, 27384951, 37261594, 38495162, 48372615, 49516273, 51627384, 59483726, 61594837, 62738495, 72615948, 73849516, 83726159, 84951627, 94837261, 95162738]
This list is not in oeis.
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def isrobust(n):
""" Robust is 8 digits, no zeros, diffs all 4 or 5 """
adjacents = ['', '56', '67', '78', '89', '19', '12', '23', '34', '45']
s = str(n)
if len(s) != 8:
return False
if '0' in s:
return False
for i,v in enumerate(s):
if i == 0:
continue
if v not in adjacents[int(s[i-1])]:
return False
return True
rob = 0
sup = 0
bigsum = 0
superrobusts = []
for n in range(10000000, 100000000):
s = str(n)
if isrobust(n):
rob += 1
if len(set(s)) == len(s):
sup += 1
bigsum += n
superrobusts.append(n)
print(rob, sup)
print(bigsum)
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Posted by Larry
on 2024-06-02 13:18:30 |
Solution: Analytical with Computer confirmation
