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Solution:
1522343
4657981
584x187
1231842
3318768
Alan Lemm
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jdsmith sent a picture of a solution.
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Solve for "SIMPLE ADDITION, by John Gowland"
A=152
B=234
C=657
D=981
E=584
F=187
G=123
H=184
J=318
K=768
a=14
b=568
c=27
d=48
e=317
f=543
g=918
h=513
j=846
k=23
m=18
n=28
Marcis Petersons
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1522343
4657981
584x187
1231842
3318768
Triads
152,784,936
196,542,738
146,583,729
317,529,846
184,392,576
186,543,729
192,384,576
342,576,918
324,657,981
Dwight Kidder
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Hello!
Solution to XmasPuzzle:
1522343
4657981
584.187
1231842
3318768
First I thought it would be impossible by hand, but
the key for me was: K/4 + K/2 = (d/2)^2
Best Xmas, Juha
-------------------------------------------------------1522343
4657981
584 187
1231842
2318768
I didn't understand the problem for the longest time. I assumed x,y,z were
fixed. Once I understood it, it became easier...
I assume the rubric means that each row in the table consists of a sequence
of 3 numbers, the last of which is the sum of the first 2, taking all digits
1-9 once and never taking the digit 0. Then it may seem that there are three
ways to fit the numbers together:
???+???=??? (1)
??+???=???? (2)
?+????=???? (3)
However, (3) provides no solutions. The last three digits of the left-hand
4-digit number must be at most 987. The 1-digit number must be at most 9.
Adding these together, we get at most 996, and are unable to carry anything
over to give different first digits in the 2 4-digit numbers.
Similarly, (2) provides no solutions, as the left hand side is less than
100+1000=1100, and so the right hand side must have a 0 digit in it
somewhere. Therefore all solutions are of form (1).
Let x+y=z be such that x,y,z are 3-digit numbers which between them include
all the digits 1-9. Then x+y+z is a multiple of 9, but x+y+z=2z. Therefore z
is a multiple of 9. Also z>=135+246=381 and z has 3 different digits, none
of which is 0.
Going through all such z, finding valid x,y (noting that each row contains a
square) we find that each triple is either in the following sequence or the
sequence you get by swapping the first two terms of each element:
[ 194, 382, 576 ],
[ 192, 384, 576 ],
[ 184, 392, 576 ],
[ 182, 394, 576 ],
[ 186, 543, 729 ],
[ 183, 546, 729 ],
[ 146, 583, 729 ],
[ 143, 586, 729 ],
[ 196, 542, 738 ],
[ 243, 576, 819 ],
[ 317, 529, 846 ],
[ 135, 729, 864 ],
[ 324, 567, 891 ],
[ 342, 576, 918 ],
[ 152, 784, 936 ],
[ 324, 657, 981 ]
We now take each row of the matrix, one at a time.
3) E/4,11J/6,c^2
E is 3-digits long, and therefore the smallest number in the triple must be
less than 250. The second number must be a multiple of 11. The largest
number is a square and so 576 or 729.
Therefore the triple is
146,583,729: E=584,J=318,c=27
...2...
...7...
584 ...
.......
.318...
5) H,2a^2,(d/2)^2
The second number is twice a square and the largest number is a square (576
or 729).
Therefore the triple is
184,392,576:H=184,a=14,d=48
1..2.4.
4..7.8.
584 ...
...184.
.318...
7) K/4,K/2,(d/2)^2
The second number is twice the first.
Therefore the triple is
192,384,576:K=768,d=48
1..2.4.
4..7.8.
584 ...
...184.
.318768
1) A,n^2,4*B
The largest number must be a multiple of 4 and therefore either 576,864 or
936. Then given that the second number must be a square, we have that the
triple is either
135,729,864: A=135, n=27, B=216
or
152,784,936: A=152, n=28, B=234
Furthermore, looking at the grid, n ends with 8. Therefore A=152,n=28,B=234
152234.
4..7.8.
584 ...
...1842
.318768
2) a^2,b-B/9,6*G
The largest number must be a multiple of 6 and therefore either
576,738,846,864,918,938. Given that the first number must be 14^2=196, the
triple is
196,542,738: a=14, b-B/9=542, G=123
Since B=234, B/9=26 and therefore b=568.
152234.
46.7.8.
584 ...
1231842
.318768
4) e,k^2,j
Note from the grid that k=23. Therefore, the triple is
317,529,846:e=317,k=23,j=846
1522343
46.7.81
584 .87
1231842
.318768
9) m^2,C,D
From the grid, we have m=18, C=6?7, D=?81. Solving 324+6?7=?81 gives
324+657=981
Therefore C=657,D=981.
1522343
4657981
584 .87
1231842
.318768
6) h-D/3,f,c^2
From the grid, f=543, c^2=27^2=729. Now 729-543=186. Therefore h-D/3=186.
and h=186+D/3=186+(981/3)=186+327=513
1522343
4657981
584 .87
1231842
3318768
8) k^2-F,(d/2)^2,g
Letting the missing digit be x, we see from the grid that
k^2-F=529-(100x+87)=442-100x. (d/2)^2=576. g=908+10x.
Therefore 1018-100x=908+10x. Solving for x, we find 110=110x, and x=1.
Therefore k^2-F=342, (d/2)^2=576, g=918, and the grid is
1522343
4657981
584 187
1231842
3318768
Luke Pebody
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1522343
4657981
584X187
1231842
3318768
Sorry about the lack of separators, but typing in text-only does that to
you. Good problem, and I love the fiendish way the clues end up
interacting with each other. I've never solved one of these before,
either. Cheers to you all!
Darrel C Jones