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Prime Square Dice (Posted on 2011-09-10) Difficulty: 3 of 5
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

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  Subject Author Date
SolutionsolutionDej Mar2011-09-10 13:48:40
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