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Happy 2009 to you all (Posted on 2009-02-08) Difficulty: 3 of 5
Solve this crossnumber puzzle:

A B C D E
F G
H I J K L
M N
O P
Q R
S T U V W
X
Y Z
      

         D*V       = 2009 	    N + T + U   = 2009
         F2 – C    = 2009 	    O – L3      = 2009
         G + H     = 2009 	    P – O       = 2009
         I / X     = 2009	    3Q – A      = 2009
         J/2 – Z/3 = 2009           R2 – B      = 2009
         K – E     = 2009 	    Y2 + 2N – W = 2009
         M + S/4   = 2009
No number begins with zero.

Note: This nice and ingenious puzzle (look at the symmetry of the grid) and some similar ones are © copyrighted but his author has granted permission for its use.
His site (with many interesting puzzles) I´ll mention in the official solution.

See The Solution Submitted by pcbouhid    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
solution (no explanation) | Comment 2 of 4 |
1920#5922
##4#2#359
1#19643#3
750#3#0##
6984#1248
3##0###12
46#2856#5
A=241 B=1472 C=295
D=49 E=91293 F=48
G=89 H=1920 I=24108
J=5922 K=93302 L=25
M=263 N=359 O=17634
P=19643 Q=750 R=59
S=6984 T=402 U=1248
V=41 W=825 X=12
Y=46 Z=2856
if the logic I used to get this solution is
correct then this is the only solution.  My
method of solving was a combination of spread
sheets and qbasic programming.  I did not make
one large program to solve the whole thing but
simply made individual programs as I went along
to help elimiate various digit combinations for
each number and as I did so I was able to deduce
each number in turn.  What I started with was
D*V=2009 since 2009=7^2 * 41 and 41 is prime then
the only way 2009 can be the multiplication of 2 2-digit
numbers is if they are 41 and 49 in some order.  I then
used the fact that I/X=2009 and if V=49 then X would be
at least 90 and thus would cause I to have more than 5 digits.
Thus V=41 and D=49.  And everything else followed from that.

  Posted by Daniel on 2009-02-09 00:34:54
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