 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Prime Square Dice (Posted on 2011-09-10) I have in front of me five standard dice, touching in a row, left to right.

Treating each face as a base-10 digit with the value of its number of pips, the five dice, on their faces that are toward me, form a 5-digit prime. The tops of the dice, also taken in order, form a 5-digit perfect square.

If I were looking at these five dice from the opposite direction, I'd see a different prime number formed by the digits on the vertical faces in order, and I'd see a different perfect square formed by the tops, again taken in order, as seen. In fact, that perfect square would be larger than the one I'm seeing from the side I'm actually on.

And one more thing: the five digits of each prime are different, but of course any given digit might or might not be on both primes.

1. Identify the primes and squares involved.

2. If duplicate digits (any multiples of the same digit within a number) were allowed, what could the front and back primes be, and the resulting squares on top, other than the ones found in part 1, using the same other rules as part 1?

 Submitted by Charlie Rating: 4.0000 (1 votes) Solution: (Hide) I'm seeing 43651 on the front facing me, with 12544 as the square on top. The prime on back is 62143, and the square that's the reverse of the first is 44521. The program does not check for duplicate digits within the primes and so also reports the part 2 solution: 43411 as the original prime and 66343 on back, with the squares on top the same as in the first solution. ``` 5 cls 10 for Sr=100 to 316 20 Sq=cutspc(str(Sr*Sr)) 30 Sqrev="" 40 Good=1 50 for I=1 to 5 60 Sqrev=mid(Sq,I,1)+Sqrev 70 if instr("0789",mid(Sq,I,1))>0 then Good=0 80 next I 90 Sq2=val(Sqrev) 100 Sr2=int(sqrt(abs(Sq2))+0.5) 110 if Sr2*Sr2=Sq2 and Good=1 and Sq2>val(Sq) then 210 :for T1=1 to 6 220 :if T1<>val(mid(Sq,1,1)) and T1<>7-val(mid(Sq,1,1)) then 230 :for T2=1 to 6 240 :if T2<>val(mid(Sq,2,1)) and T2<>7-val(mid(Sq,2,1)) then 250 :for T3=1 to 6 260 :if T3<>val(mid(Sq,3,1)) and T3<>7-val(mid(Sq,3,1)) then 270 :for T4=1 to 6 280 :if T4<>val(mid(Sq,4,1)) and T4<>7-val(mid(Sq,4,1)) then 290 :for T5=1 to 6 300 :if T5<>val(mid(Sq,5,1)) and T5<>7-val(mid(Sq,5,1)) then 310 :Pr1s=cutspc(str(T1))+cutspc(str(T2))+cutspc(str(T3))+cutspc( str(T4))+cutspc(str(T5)) 320 :Pr2s=cutspc(str(7-T5))+cutspc(str(7-T4))+cutspc(str(7-T3))+c utspc(str(7-T2))+cutspc(str(7-T1)) 330 :Pr1=val(Pr1s):Pr2=val(Pr2s) 340 :if prmdiv(Pr1)=Pr1 and prmdiv(Pr2)=Pr2 then 350 :print Pr2:print Sq2:print Sq:print Pr1:print 360 :endif 370 :endif 380 :next T5 390 :endif 400 :next T4 410 :endif 420 :next T3 430 :endif 440 :next T2 450 :endif 460 :next T1 920 :endif 930 next Sr ``` Subject Author Date solution Dej Mar 2011-09-10 13:48:40 Please log in:

 Search: Search body:
Forums (0)