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Killer-X-tra (2) (Posted on 2008-06-20) Difficulty: 4 of 5
Place the numbers 1 to 9 once in each row, column, long diagonal and 3x3 box. Obviously, the central yellow-coloured 3x3 box is common to both grids.

The number in the cage (or shaded area) represents the result of an arithmetical operation by its accompanying sign. For example, 4÷ in a two-celled cage means that it contains either 8 and 2 or 4 and 1 (in any order).

Unlike Killer Sudoku, the same number may appear more than once in a cage. So, 28x in a cage which overlaps the 3x3 grids may contain two 2s and a 7.



Once again, I wish to thank brianjn most sincerely for his continued support. He converted my black and white Excel image to a coloured one in 'jpg' format.
Acknowledgement to Peter at sudexel.com for beta testing the puzzle.

No Solution Yet Submitted by Josie Faulkner    
Rating: 4.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution / (Long) Walkthrough | Comment 9 of 16 |

Third, here is the (long) walkthrough:

 

Let me start by explaining how I am going to refer to each cell.  I will name each 3x3 grid with a capital letter, starting in the top left corner and going to the right.  So they are:

 

ABC

DEF

GHIJK

  LMN

  OPQ

 

And in each grid I will name each cell with a lower case letter, again starting in the top left and going to the right.  So they are:

 

abc

def

ghi

 

So the top most row would consist of cells Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb, and Cc.

 

And we begin...

 

Grids D, G and H:

        360 = 5*(72) so either Gb, Ge or Gf is 5.

        15 = 5*(3).  Ga can't be 5 due to the 5 in either Gb, Ge or Gf.  So Dg=5 and Ga=3.

        70 = 5*(14).  Di can't be 5 due to the 5 in Dg.  Gc can't be 5 due to the 5 in either Gb, Ge or Gf.  So Ha=5.

        42 = 7*(6) so either Gd or Gg is 7.

        70 = 7*(10).  Gc can't be 7 due to the 7 in either Gd or Gg.  So Di=7 and Gc=2.

        4 = 1+3.  Gi can't be 3 due to the 3 in Ga.  So Hg=3 and Gi=1.

        42 = 6*7 (Gd and Gg), and 360 = 5*8*9 (Gb, Ge and Gf).  So for grid G to have a 4 it must be that Gh=4.

 

Grids A, B and C:

        127 = 7*(18).  Aa and Ad can't be 7 due to the 7 in either Gd or Gg.  So Ab=7.

        224 = 7*(32).  Ai and Dc can't be 7 due to the 7 in Di.  So Bg=7.

        15 = 5*(3).  We need a third factor so we must have 15 = 5*3*1.  Ba can't be 5 due to the 5 in Ha, and it can't be 3 due to the 3 in Hg.  So Ba=1.

        56 = 7*(8).  Cc can't be 7 due to the 7 in Ab.  So Cf=7 and Cc=8.

 

Grids L, M, O and P:

        378 = 7*(54) so either Ob, Oe or Of is 7.

        21 = 7*(3).  Oa can't be 7 due to the 7 in either Ob, Oe or Of.  So Lg=7 and Oa=3.

        5 = 5/1 is the only way to have this division with digits <= 9.  So Md and Mg are 1 and 5.

        6 = 1+5 = 2+4.  Pg can't be 1 or 5 due to the 1 and 5 in Md and Mg.  So Oi and Pg are 2 and 4.

        8 = 8/2 = 4/1 are the only ways to have this division with digits <= 9.  Either way, Od and Og can't be 5.

        378 = 7*6*9 (Ob, Oe and Of), and 288 = 8*6*6 = 8*4*9 (Oc).  So for grid O to have a 5 it must be that Oh=5.

        378 = 7*6*9 (Ob, Oe and Of), 288 = 8*6*6 = 8*4*9 (Oc), and 6 = 2+4 (Oi).  So for grid O to have a 1 it must be that Od or Og is 1, which means Od and Og are 1 and 4.  Og can't be 4 since either Oi or Pg is 4.  So Od=4 and Og=1.

        6 = 2+4.  Oi can't be 4 due to the 4 in Od.  So Pg=4 and Oi=2.

        378 = 7*6*9 (Ob, Oe and Of).  So for grid O to have an 8 it must be that Oc=8.

 

Grids J, K and N:

        180 = 5*(36) so either Kd, Ke or Kh is 5.

        40 = 5*(8).  Ka can't be 5 due to the 5 in either Kd, Ke or Kh. So Jc=5 abd Ka=8.

        243 = 3^5.  The only way this can be arranged into three factors <= 9 is if we have 243 = 9*9*3.  We can't have two 9s in the same row or column so we must have Kg=3, Ji=9 and Na=9.

        180 = 5*(36) = 5*6*6 = 5*4*9.  We can't have two 6s in the same grid so we must have 180 = 5*4*9.  Kd and Kh can't be 9 due to the 9s in Ji and Na.  So Ke=9.

 

Grids P and Q:

        360 = 5*(72) = 5*8*9.  Qh and Qi can't be 5 due to the 5 in Oh.  So Qf=5.  Qh can't be 9 due to the 9 in Ke.  So Qi=9 and Qh=8.

        The 5 in grid P can't be in the left column due to the 5 in either Md or Mg.  5 can't be in the right column due to the 5 in Jc.  5 can't be in the middle or bottom row due to the 5s in Qf and Oh.  So Pb=5.

 

Grids I, M and Q (long diagonal):<o:p></o:p>

        Qa, Qe and Qi can't be 5 due to the 5 in Qf.

        Ma, Me and Mi can't be 5 due to the 5 in either Md or Md.

        Ia cant' be 5 due to the 5 in Ha (and Jc).  Ie can't be 5 due to the 5 in Oh.  So Ii=5.

 

Grid K:

        180 = 5*4*9.  Kh can't be 5 due to the 5 in Ii.  So Kd=5 and Kh=4.

 

Grids K, M and O (long / diagonal):

        Kc, Ke and Kg can't be 5 due to the 5 in Kd.

        Oc, Oe and Og can't be 5 due to the 5 in Oh.

        Mc can't be 5 due to the 5 in Jc.  Me can't be 5 due to the 5 in Pb.  So Mg=5 and Md=1.

 

Grids I, L, O and P:

        The 7 in grid I can't be in the left or right columns due to the 7s in Lg and Cf.  So Ib, Id or Ih is 7.

        378 = 7*6*9.  Ob and Oe can't be 7 due to the 7 in Ib, Id or Ih.  So Of=7.  Oe can't be 9 due to the 9 in Ke (diagonal rule).  So Ob=9 and Oe=6.

        288 = 8*6*6 = 8*4*9.  Pa can't be 4 or 9 due to the 4 in Pg and the 9 in Ob.  So we must have 288 = 8*6*6.  So Pa=6 and Li=6.

 

Grid Q:

        6 can't be in the top or middle rows due to the 6s in Pa and Oe.  So Qg=9.

        3 can't be in the top row due to the 3 in Oa and 3 can't be in the left column due to the 3 in Kg.  So Qe=3.

 

Grid M:

        9 can't be in the diagonal due to the 9 in Qi.  9 can't be in the top row due to the 9 in Na.  And 9 can't be in the right column due to the 9 in Ji.  So Mh=9.

 

Grid I:

        9 can't be in the diagonal due to the 9 in Qi.  9 can't be in the middle or bottom rows due to the 9s in Ke and Ji.  9 can't be in the middle column due to the 9 in Ob.  So Ic=9.

 

Grid L:

        9 can't be in the middle or right columns due to the 9s in Ob and Ic.  9 can't be in the top row due to the 9 in Na.  So Ld=9.

        5 can't be in the middle or right columns due to the 5s in Oh and Ii.  So La=5.

        14 = 5+(9).  Since grid L already has 9, 7, 6, and 5 placed, the only way to have two digits add up to 9 is if they are 1 and 8.  Lc can't be 8 due to the 8 in Oc.  So Lb=8 and Lc=1.

 

Grids A, E and I (long diagonal):

        Aa, Ae and Ai can't be 7 due to the 7 in Ab.

        In grid J, 56 = 7*(8) so either Jd, Je or Jf is 7.

        Ia, Ie and Ii can't be 7 due to the 7s in Lg, Of and either Jd, Je or Jf.

        Ea can't be 7 due to the 7 in Bg.  Ei can't be 7 due to the 7 in Di.  So Ee=7.

 

Grid G:

        360 = 5*8*9.  Gb can't be 5 or 9 due to the 5 in Ha and the 9 in Ic.  So Gb=8.

        42 = 7*6.  Gg can't be 7 due to the 7 in Ee (diagonal rule).  So Gd=7 and Gg=6.

 

Grid E:

        24 = (2^3)*3 = 8*3 = 4*6.  Ed and Eg can't be 3 due to the 3 in Hg.  So it must be that 24 = 4*6.  Eg can't be 6 due to the 3 in Gg (diagonal rule).  So Ed=6 and Eg=4.

 

Grid I:

        The top and bottom row can't have a 3 due to the 3s in Ga and Hg.  The left column can't have a 3 due to the 3 in Oa.  3 can't be in the diagonal due to the 3 in Qe.  So If=3.

        Consider the column that starts at Ia and ends at Og.  Ia can only be 2, 6 or 8 due to the numbers placed in this column so far.  Ia can't be 2 or 8 due to the 2 in Gc and the 8 in Gb.  So Ia=6.

 

Grid J:

        56 = 7*8 = 7*8*1 = 8*4*2.  Either way, Jd, Jd and Jf can't be 6.

        The top row can't have a 6 due to the 6 in Ia.  The left column can't have a 6 due to the 6 in Pa.  So Jh=6.

 

Grid K:

        The top and bottom rows can't have a 6 due to the 6s in Ia and Jh.  So Kf=6.

 

Grid M:

        6 can't be in either or / diagonal due to the 6s in Ia and Oe.  The middle column can't have a 6 due to the 6 in Jh.  So Mf=6.

        The top row can't have an 8 due to the 8 in Lb.  8 can't be in the / diagonal due to the 8 in Oc.  So Mi=8.

        3 can't be in either or / diagonal due to the 3s in Qe and Kg.  So Mb=3.

 

Grid N:

        The left and right columns can't have a 6 due to the 6s in Qg and Kf.  The middle and bottom rows can't have a 6 due to the 6s in Mf and Li.  So Nb=6.

        The left and right columns can't have a 5 due to the 5s in Kd and Qf.  The bottom row can't have a 5 due to the 5 in Mg.  So Ne=5.

        The top and bottom rows can't have an 8 due to the 8s in Lb and Mi.  The left column can't have an 8 due to the 8 in Ka.  So Nf=8.

        The left and middle columns can't have a 3 due to the 3s in Kg and Qe.  The top row can't have a 3 due to the 3 in Mb.  So Ni=3.

 

Grid P:

        The top and middle rows can't have a 3 due to the 3s in Oa and Qe.  The middle column can't have a 3 due to the 3 in Mb.  So Pi=3 (how fitting!).

        Consider the row that starts at Og and ends at Qi.  Ph can only be 7 due to the numbers already placed in this row so far.  So Ph=7.

        The middle and right columns can't have a 9 due to the 9s in Mh and Ji.  So Pd=9.

        The right column can't have an 8 due to the 8 in Mi.  So Pe=8.

 

Grid L:

        The bottom row can't have 3 due to the 3 in Ni.  The right column can't have a 3 due to the 3 in If.  So Le=3.

        The right column can't have a 2 due to the 2 in Oi.  So Lh=2.

        The only remaining number is 4 so Lf=4.

 

Grid M:

        The left column can't have a 4 due to the 4 in Pg.  The middle row can't have a 4 due to the 4 in Lf.  So Mc=4.

        The middle column can't have a 7 due to the 7 in Ph.  So Ma=7.

        The only remaining number is 2 so Me=2.

 

Grid N:

        The top and bottom rows can't have a 7 due to the 7s in Ma and Lg.  So Nd=7.

        The bottom row can't have a 2 due to the 2 in Lh.  So Nc=2.

        The middle column can't have a 4 due to the 4 in Kh.  So Ng=4.

        The only remaining number is 1 so Nh=1.

 

Grid K:<o:p></o:p>

        The right column canít have a 2 due to the 2 in Nc.  So Kb=2.<o:p></o:p>

        The / diagonal canít have a 1 due to the 1 in Og.  So Ki=1.<o:p></o:p>

        The only remaining number is 7 so Kc=7.<o:p></o:p>

 <o:p></o:p>

Grid Q:<o:p></o:p>

        The left and right columns canít have a 7 due to the 7s in Nd and Kc.  So Qb=7.<o:p></o:p>

        The left column canít have a 4 due to the 4 in Ng.  So Qc=4.<o:p></o:p>

        The \ diagonal canít have a 2 due to the 2 in Me.  So Qd=2.<o:p></o:p>

        The only remaining number is 1 so Qa=1.<o:p></o:p>

 <o:p></o:p>

Grid P:<o:p></o:p>

        The top row canít have a 1 due to the 1 in Qa.

  Posted by nikki on 2008-06-24 01:54:11
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