List all perfect square numbers over 1000, employing only 2 different decimal digits.
Example: 88^2=7744 will be on your list.
Ignore trivial solutions like 1 or 4 or 9 followed by an even number of zeroes.
How far can you go?
Known largest solution is
a 10-digit number 6,661,661,161.
I tried to get past 18 digits (where integer*8 ran out) by making a "squaring' routine in software. By execution time proved intolerable.
So here is the hardware square version...
rabbit-3:~ lord$ t30
winner 16 4
winner 25 5
winner 36 6
winner 49 7
winner 64 8
winner 81 9
winner 121 11
winner 144 12
winner 225 15
winner 441 21
winner 484 22
winner 676 26
winner 1444 38
winner 7744 88
winner 11881 109
winner 29929 173
winner 44944 212
winner 55225 235
winner 69696 264
winner 9696996 3114
winner 6661661161 81619
C^C
rabbit-3:~ lord$
rabbit-3:~ lord$
rabbit-3:~ lord$
rabbit-3:~ lord$ more t30.f
program testtt
integer*8 i,dum,d,ten,n
integer flag,dtime,list(2),ndig,num(18)
ten=10
do i = 4,999999999
list(2)=-1
dum=i*i
ndig=1+log10(1.*dum)
if(ndig.gt.18)stop
d=dum
do n=ndig-1,0,-1
p=ten**n
num(n+1)=dum/ten**n
dum = dum - num(n+1)* ten**n
enddo
if(num(1).eq.0) go to 2
list(1)=num(1)
do 1 k=2,ndig
if(list(2).eq.-1.and.num(k).ne.list(1))then
list(2)=num(k)
go to 1
endif
if(num(k).ne.list(1).and.num(k).ne.list(2))go to 2
1 enddo
if(num(1).ne.0)print*,'winner ',d,i
2 enddo
end
Edited on October 23, 2018, 4:45 am
different language - same result
