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.
for d=1:7
st=str2double(['2' repmat('0',1,d-1)]);
nd=str2double(repmat('8',1,d));
ct=0;
for n=st:2:nd
n3 = n * 3;
tst = [char(string(n)) char(string(n3))];
good = true;
for i = 1:length(tst)
if ~contains('02468', tst(i))
good = false;
break
end
end
if good
ct = ct + 1;
% disp([ n n3])
end
end
disp([d ct])
end
finds
number possible values
of digits of N
1 2
2 7
3 24
4 82
5 280
6 956
7 3264
Looking up the second column of numbers in the OEIS results in A003480 and A020727.
The former is described as a(n) = 4*a(n-1) - 2*a(n-2) (n >= 3) with elements 0 through 2 being 1, 2, 7. The latter sequence has a comment "It appears that a(n) = 4*a(n-1) - 2*a(n-2) (holds at least up to n = 1000 but is not known to hold in general)". But we're concerned only as far as 22, so they agree up to where we need one of them.
N(1)=2; N(2)=7;N(3)=24;
for i=4:25
N(i)=4*N(i-1) - 2*N(i-2);
fprintf('%3d %14d\n',i, N(i));
end
produces
4 82
5 280
6 956
7 3264
8 11144
9 38048
10 129904
11 443520
12 1514272
13 5170048
14 17651648
15 60266496
16 205762688
17 702517760
18 2398545664
19 8189147136
20 27959497216
21 95459694592
22 325919783936
23 1112759746560
24 3799199418368
25 12971278180352
So, for 22-digit numbers, N has 325,919,783,936 possible values.
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Posted by Charlie
on 2023-07-29 12:13:01 |
same answer as Larry's
