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Clueless (Posted on 2005-03-14) Difficulty: 3 of 5
I gave my niece and nephew the following puzzler:

A: Draw a 3x3 grid, with the boxes labeled 1 through 9 in the usual order (left to right; top to bottom).

B:For each of the following instructions you must write a number, greater than 10, starting in one box and going across (left to right) or down, with one digit to each box. Your answers should fill the grid, with no two answers overlapping.

C:Starting with a box whose number is a square, write a square number.

D:Starting with a box whose number is a cube, write a cube number.

E:Starting in a box whose number is prime, write a prime number whose digits add up to an even number.

F:Starting with a box whose number is even, write an even number

Unfortunately, one of the copies of my instructions had part F completely missing. Each child turned in a 3x3 grid which was correct for the version they had been given. By coincidence, they turned in identical grids.

How did they fill out their grids?

See The Solution Submitted by Sam    
Rating: 4.0000 (1 votes)

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Solution Computer search -- spoiler | Comment 3 of 6 |

The following program tests all sets of 3-digit cube in the first row, square in the second row and prime in the third, against each of the 4-clue formats mentioned in my previous post.

    5   open "cluels2.txt" for output as #2
   10   for CBase=5 to 9
   20   Cube=CBase*CBase*CBase
   30   for SBase=10 to 31
   40   Square=SBase*SBase
   50   Prime=2
   60   while Prime<100
   70        Prime=nxtprm(Prime)
   80   wend
   90   while Prime<1000
  100     A$=cutspc(str(Cube))
  110     B$=cutspc(str(Square))
  120     C$=cutspc(str(Prime))
  130     VSq=val(left(A$,1)+left(B$,1)+left(C$,1))
  135     Sr=int(sqrt(VSq)+0.5)
  140     Even=val(mid(A$,2,2))
  150     Pr2=val(mid(B$,2,2))
  160     Cu2=val(mid(C$,2,2))
  165     Cr=int(Cu2^(1/3)+0.5)
  168     if Pr2>1 then
  170     :if Even@2=0 and nxtprm(Pr2-1)=Pr2 and Cr*Cr*Cr=Cu2 then
  175       :if Sr*Sr=VSq then
  179          :print#2, "Vert Sq 3"
  180         :print#2, A$:print#2, B$:print#2, C$:print#2,
  230     VSq=val(left(B$,1)+left(C$,1))
  235     Sr=int(sqrt(VSq)+0.5)
  240     Pr2=val(mid(B$,2,1)+mid(C$,2,1))
  250     Even=val(mid(B$,3,1)+mid(C$,3,1))
  268     if Pr2>1 then
  270     :if Even@2=0 and nxtprm(Pr2-1)=Pr2 then
  275       :if Sr*Sr=VSq then
  279          :print#2, "Hor cube (orig)"
  280         :print#2, A$:print#2, B$:print#2, C$:print#2,
  330     VSq=val(left(A$,1)+left(B$,1)+left(C$,1))
  335     Sr=int(sqrt(VSq)+0.5)
  340     Even=val(mid(A$,2,1)+mid(B$,2,1))
  350     Pr2=val(mid(A$,3,1)+mid(B$,3,1))
  360     Cu2=val(mid(C$,2,2))
  365     Cr=int(Cu2^(1/3)+0.5)
  368     if Pr2>1 then
  370     :if Even@2=0 and nxtprm(Pr2-1)=Pr2 and Cr*Cr*Cr=Cu2 then
  375       :if Sr*Sr=VSq then
  379          :print#2, "Vert Sq 3*"
  380         :print#2, A$:print#2, B$:print#2, C$:print#2,
  430     VSq=val(left(B$,1)+left(C$,1))
  435     Sr=int(sqrt(VSq)+0.5)
  440     Pr2=val(mid(B$,2,2))
  450     Even=val(mid(C$,2,2))
  468     if Pr2>1 then
  470     :if Even@2=0 and nxtprm(Pr2-1)=Pr2 then
  475       :if Sr*Sr=VSq then
  479          :print#2, "Hor cube (orig)*"
  480         :print#2, A$:print#2, B$:print#2, C$:print#2,
  530     HSq=val(A$)
  535     Sr=int(sqrt(HSq)+0.5)
  540     Pr2=val(mid(B$,2,2))
  550     Even=val(mid(B$,1,1)+mid(C$,1,1))
  560     Cu2=val(mid(C$,2,2))
  565     Cr=int(Cu2^(1/3)+0.5)
  568     if Pr2>1 then
  570     :if Even@2=0 and nxtprm(Pr2-1)=Pr2 then
  575       :if Sr*Sr=HSq and Cr*Cr*Cr=Cu2 then
  579          :print#2, "Hor square (orig cube)"
  580         :print#2, A$:print#2, B$:print#2, C$:print#2,
 1295     Prime=nxtprm(Prime)
 1300   wend
 1310   next
 1320   next

The check did not include that the prime numbers consist of digits that add up to an even number, and so produced several more results than just those that meet the full criteria.  But inspection of the output shows two cases where the prime numbers' digits each added up to an even number:

Vert Sq 3*
125
289
101
Vert Sq 3*
125
289
127

In the first instance, the 2-digit cube beginning in square 8 has a leading zero, and is presumed disallowed.  That leaves the second one shown.

So one of the children had 125 as the cube, 289 as the square and 127 as the prime.  The other had 121 as the square, 28 as the even number, 59 as the prime and 27 as the cube.


  Posted by Charlie on 2005-03-14 20:49:54
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