Six of the factors of 1221 are palindromes: {1,3,11,33,111,1221} but two are not: {37,407}
(A) What number under one million has the most palindrome factors?
Some numbers have only palindromes as factors. Examples include 88: {1,2,4,8,11,22,44,88} and any palindromic prime
(B) Find the smallest number with at least as many palindrome factors as part (A) but having only palindrome factors.
Part 1:
max=[0 0]; clc
for n=1:1000000
k=1:ceil(sqrt(n));
f=k(rem(n,k)==0);
f2=n./f;
if f(end)==f2(end)
f=[f f2(1:end-1)];
else
f=[f f2];
end
ct=0;
for i=1:length(f)
if isPalin(f(i))
ct=ct+1;
end
end
% disp([n ct])
if ct>max(end,2)
max=[max;[n ct]];
disp([n ct length(f)])
for i=1:length(f)
if isPalin(f(i))
fprintf('%4d ',f(i))
end
end
disp(' ')
end
end
disp(max)
function ip=isPalin(x)
s=char(string(x));
bynd=length(s)+1;
ip=true;
for i=1:floor(length(s)/2)
if s(i)~=s(bynd-i)
ip=false;
break
end
end
end
finds
792792 39 144
1 2 3 4 6 7 8 9 11 22 33 44 66 77 88 99 121 242 252 363 484 616 858 99099 88088 66066 44044 33033 22022 11011 9009 8008 6776 6006 4004 3003 2772 2002 1001
indicating 39 palindromic divisors of 792792 out of a total of 144 divisors. The palindromic divisors are then listed.
Part 2:
max=[0 0]; clc
for n0=1:100000
ns=char(string(n0));
ns1=ns;
for i=length(ns):-1:1
ns1=[ns1 ns(i)] ;
end
n1=str2double(ns1);
ns2=ns;
for i=length(ns)-1:-1:1
ns2=[ns2 ns(i)] ;
end
n2=str2double(ns2);
for n=[n1 n2]
k=1:ceil(sqrt(n));
f=k(rem(n,k)==0);
f2=n./f;
if f(end)==f2(end)
f=[f f2(1:end-1)];
else
f=[f f2];
end
ct=0; good=true;
for i=1:length(f)
if isPalin(f(i))
ct=ct+1;
else
good=false;
end
end
if good==false
continue
end
% disp([n ct])
if ct>max(end,2)
max=[max;[n ct]];
disp([n ct length(f)])
for i=1:length(f)
if isPalin(f(i))
fprintf('%4d ',f(i))
end
end
disp(' ')
end
end
end
% disp(max)
function ip=isPalin(x)
s=char(string(x));
bynd=length(s)+1;
ip=true;
for i=1:floor(length(s)/2)
if s(i)~=s(bynd-i)
ip=false;
break
end
end
end
Finds only as many as 18 such divisors:
448844 18 18
1 2 4 11 22 44 101 202 404 448844 224422 112211 40804 20402 10201 4444 2222 1111
having tested through 10-digit palindromes, as one of the palindromic divisors needs to be the number itself. Trying for more digits would take much longer to run.
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Posted by Charlie
on 2022-03-15 09:22:53 |
part (A) solution and an attempt at (B)
