N is a number with the property that all its digits are even.
3N also has that property.
Example: N=208 and 3N=624.
How many possible values of N have 22 digits? (3N may have 22 or 23 digits.)
For reference:
Evened N and 3N is a version of this problem where N is only a 3 digit number.
Working this out for number of digits 0 to 6 gives:
1, 2, 7, 24, 82, 280, 956
Some observations:
Any N between (1/3) and (2/3) of a power of 10 will result in 3N being one digit longer and starting with a 1.
In fact N can begin only with {2,6,8};
N can only end with {0,2,8}
digits other than the first and last are restricted to {0,2,6,8}
a 6 can be followed only by {6,8}
I suspect these digit restrictions may be enough to find a solution but I have not tested it yet for numbers longer than about 6-digits.
And then there is Sloane's OEIS. Indeed the sequence shows up with two different generating rules. For readability assume somewhere mid sequence is: ..., a, b, c, ....
A003480 c = 4b-2a
A020727 c = round(b^2/a)
(each valid only after the first few elements)
The ratio of a(n)/a(n-1) approaches 2+√2
The function seqP(n), see below, generates a list for i from 1 to n, using the formula c = round(b^2/a)
Following the pattern, thanks to oeis, the count for 22-digit numbers would appear to be:
325919783936
I did make some code to directly find this for a given number of digits, but my code, in Python, on my laptop was too slow.
So I would conclude that 325919783936 is probably the answer, but not confirmed.
------------
def isAllEven(n):
""" input an integer, if all digits are even then output True """
odds = '13579'
if n >= 10:
return isAllEven(n//10) and isAllEven(n%10)
if n < 10 and str(n) in odds:
return False
if n < 10 and str(n) not in odds:
return True
def seqP(n):
ans = [1,2,7]
for i in range(3,n+1):
ans.append(round(ans[i-1]**2 / ans[i-2]))
return ans
------------
# This code will give solution for a given number of digits, but gets slow for larger numbers
ns = []
count = 0
numDigits = 4
for i in range(10**(numDigits-1),10**(numDigits),2):
if isAllEven(i):
if isAllEven(3*i):
count += 1
# print(i,3*i)
ns.append(i)
print(len(ns))
------------
# This code will get solution for one more number of digits
sols = [[0],[2,8],[20, 22, 28, 68, 80, 82, 88],[200, 202, 208, 220, 222, 228, 268, 280, 282, 288, 668, 680, 682, 688, 800, 802, 808, 820, 822, 828, 868, 880, 882, 888]]
sizes = [len(i) for i in sols]
digits = '0268'
after6 = [6,8]
endings = [0,2,8]
# take prior number of digits, try all possibles adding a digit at beginning, any gap or end.
nextlist = []
leng = len(str(sols[-1][0])) + 1
for i in sols[-1]:
s = str(i)
for d in digits:
for gap in range(leng):
if gap==0 and d==0:
continue
t = s[:gap] + d + s[gap:]
# print(s,d,gap, s[:gap] , d , s[gap:] , t)
if t[0]=='0' or t[-1]=='6' or '60' in t or '62' in t or t in nextlist:
continue
nextlist.append(t)
sols.append([int(n) for n in sorted(nextlist)])
sizes = [len(i) for i in sols]
print(sizes)
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Posted by Larry
on 2023-07-29 12:06:15 |
Probably solution, with a little help
