Enchanted Square...a maths problem
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hi...any maths wizards here??!!! please help me out with this problem! i have to hand this assignment up on wednesday!! >"<
Enchanted Square
Fill the nine cells of this 3x3 square with the numbers from 1 to 9 so that the sum of the numbers belonging to any given 2x2 square is always the same.
1 _ _
_ _ 7
_ _ _
urm..the blanks are for you to fill up the numbers with.
please don't just tell me what the numbers are...if i'm not wrong, there are 88 answers to this problem. i need the solution....how you get the numbers...best if there are equations or what.
happy trying clever ppl! ^^ -
maybe u shd buy the sudoku game and play
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this is the sudoku game??? really???
i've played the sudoku game once before on a newspaper...but that one was easy...and the newspaper said it has modified it a little...so well..don't know same or not...
please help??? >"<
i used algebra to solve the problem..but the answers are not very consistent? just wondering if there are better methods... -
my maths sux..Originally posted by applesweetgirl:this is the sudoku game??? really???
i've played the sudoku game once before on a newspaper...but that one was easy...and the newspaper said it has modified it a little...so well..don't know same or not...
please help??? >"<
i used algebra to solve the problem..but the answers are not very consistent? just wondering if there are better methods... -
1abOriginally posted by applesweetgirl:hi...any maths wizards here??!!! please help me out with this problem! i have to hand this assignment up on wednesday!! >"<
Enchanted Square
Fill the nine cells of this 3x3 square with the numbers from 1 to 9 so that the sum of the numbers belonging to any given 2x2 square is always the same.
1 _ _
_ _ 7
_ _ _
urm..the blanks are for you to fill up the numbers with.
please don't just tell me what the numbers are...if i'm not wrong, there are 88 answers to this problem. i need the solution....how you get the numbers...best if there are equations or what.
happy trying clever ppl! ^^
cd7
efg
1+7+a+b+c+d+e+f+g = 45
a+b+c+d+e+f+g = 37
1+a+c+d = a+b+d+7 = c+d+e+f = d+f+g+7
1+a+c = a+b+7 = c+e+f = f+g+7
==> 4 equations, 6 vars, unable to solve in real no. set, listing out relations:
c+1 = b+7
c+e = g+7
a+1 = e+f
a+b = f+g
==> found: b = c-6
for b,c is member of Z+ such that 1<=x<=9,
b is either 1,2,3
c is either 7,8,9
==> 7 and 1 taken
==> scenario split
scenario 1) b=2 and c=8
or
scenario 2) b=3 and c=9
SCENARIO 1 (b=2 and c=8 )
==> 1a2
==> 8d7
==> efg
a+d+9 = d+e+f+8 = d+f+g+7
==> g = e+1
==> a = e+f-1
==> 1 e+f-1 2
==> 8 d 7
==> e f e+1
==> undefined numbers 3,4,5,6,9
==> need to define d,e,f through trial and error
==> e+f-1 <= 9
==> e+f <= 10
==> e != 6 (g cannot be 7)
==> f != e+1
==> truncate e=3,f=4; e=4,f=5; e=5,f=6
==> e+f-1 != 7
==> truncate e=3,f=5; e=5,f=3
==> e+f-1 != 8
==> truncate e=3,f=6; e=5,f=4
==> listing solution(s)
e f a g d
4 6 9 5 3
4 3 6 5 9
192 162
837 897
465 435
==> 2 solutions found under SCENARIO 1
SCENARIO 2 (b=3 and c=9)
==> 1a3
==> 9d7
==> efg
a+d+10 = d+e+f+9 = d+f+g+7
==> g = e+2
==> a = e+f-1
==> 1 e+f-1 3
==> 9 d 7
==> e f e+2
==> undefined numbers 2,4,5,6,8
==> need to define d,e,f through trial and error
==> e+f-1 <= 8
==> e+f <= 9
==> e != 5 (g cannot be 7)
==> f != e+2
==> truncate e=2,f=4; e=4,f=6; e=6,f=8
==> e+f-1 != 7
==> truncate e=2,f=6, e=6,f=2
==> e+f-1 != 9
==> truncate e=2,f=8; e=8,f=2; e=4,f=6; e=6,f=4
==> listing solution(s)
e f a g d
2 5 6 4 8
4 2 5 6 8
4 5 8 6 2
163 183 153
987 927 987
254 456 426
==> 3 solutions found under SCENARIO 2
dunno if this helps. just playing around with algebra. -
wah, smlj?
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Originally posted by ^tamago^:
WTF u damn freakin pro leh
1ab
cd7
efg
1+7+a+b+c+d+e+f+g+h = 45
a+b+c+d+e+f+g+h = 37
1+a+c+d = a+b+d+7 = c+d+e+f = d+f+g+7
1+a+c = a+b+7 = c+e+f = f+g+7
==> 4 equations, 6 vars, unable to solve in real no. set, listing out relations:
c+1 = b+7
c+e = g+7
a+1 = e+f
a+b = f+g
==> found: b = c-6
for b,c is member of Z+ such that 1<=x<=9,
b is either 1,2,3
c is either 7,8,9
==> 7 and 1 taken
==> scenario split
scenario 1) b=2 and c=8
or
scenario 2) b=3 and c=9
SCENARIO 1 (b=2 and c=8 )
==> 1a2
==> 8d7
==> fgh
a+d+9 = d+g+f+8 = d+g+h+7
==> h = f+1
==> a = g+f-1
==> 1 g+f-1 2
==> 8 d 7
==> f g f+1
==> undefined numbers 3,4,5,6,9
==> need to define d,f,g through trial and error
==> g+f-1 <= 9
==> g+f <= 10
==> f != 6 (h cannot be 7)
==> g != f+1
==> truncate f=3,g=4; f=4,g=5; f=5,g=6
==> g+f-1 != 7
==> truncate f=3,g=5; f=5,g=3
==> g+f-1 != 8
==> truncate f=3,g=6; f=5,g=4
==> listing solution(s)
f g a h d
4 6 9 5 3
[b]192
837
465
==> 1 solution found under SCENARIO 1
SCENARIO 2 (b=3 and c=9)
==> 1a3
==> 9d7
==> fgh
a+d+10 = d+g+f+9 = d+g+h+7
==> h = f+2
==> a = g+f-1
==> 1 g+f-1 3
==> 9 d 7
==> f g f+2
==> undefined numbers 2,4,5,6,8
==> need to define d,f,g through trial and error
==> g+f-1 <= 9
==> g+f <= 10
==> f != 5 (h cannot be 7)
==> g != f+2
==> truncate f=2,g=4; f=4,g=6; f=6,g=8
==> g+f-1 != 7
==> truncate f=2,g=6, g=6,f=2
==> g+f-1 != 9
==> truncate f=2,g=8; f=8,g=2; f=4,g=6; f=6,g=4
==> listing solution(s)
f g a h d
2 5 6 4 8
4 5 8 6 2
163
987
254
183
927
456
==> 2 solutions found under SCENARIO 2
dunno if this helps. just playing around with algebra.
[/b]
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Wahh.... SIFU!!!
you are good!!
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Originally posted by ^tamago^:
thanks very much! although i do not really understand your scenario 2...you use maths calculator izzit? some parts i really dunno how you get... but really thanks! you are darn smart! ^^
1ab
cd7
efg
1+7+a+b+c+d+e+f+g = 45
a+b+c+d+e+f+g = 37
1+a+c+d = a+b+d+7 = c+d+e+f = d+f+g+7
1+a+c = a+b+7 = c+e+f = f+g+7
==> 4 equations, 6 vars, unable to solve in real no. set, listing out relations:
c+1 = b+7
c+e = g+7
a+1 = e+f
a+b = f+g
==> found: b = c-6
for b,c is member of Z+ such that 1<=x<=9,
b is either 1,2,3
c is either 7,8,9
==> 7 and 1 taken
==> scenario split
scenario 1) b=2 and c=8
or
scenario 2) b=3 and c=9
SCENARIO 1 (b=2 and c=8 )
==> 1a2
==> 8d7
==> efg
a+d+9 = d+e+f+8 = d+f+g+7
==> g = e+1
==> a = e+f-1
==> 1 e+f-1 2
==> 8 d 7
==> e f e+1
==> undefined numbers 3,4,5,6,9
==> need to define d,e,f through trial and error
==> e+f-1 <= 9
==> e+f <= 10
==> e != 6 (g cannot be 7)
==> f != e+1
==> truncate e=3,f=4; e=4,f=5; e=5,f=6
==> e+f-1 != 7
==> truncate e=3,f=5; e=5,f=3
==> e+f-1 != 8
==> truncate e=3,f=6; e=5,f=4
==> listing solution(s)
e f a g d
4 6 9 5 3
[b]192
837
465
==> 1 solution found under SCENARIO 1
SCENARIO 2 (b=3 and c=9)
==> 1a3
==> 9d7
==> efg
a+d+10 = d+e+f+9 = d+f+g+7
==> g = e+2
==> a = e+f-1
==> 1 e+f-1 3
==> 9 d 7
==> e f e+2
==> undefined numbers 2,4,5,6,8
==> need to define d,e,f through trial and error
==> e+f-1 <= 9
==> e+f <= 10
==> e != 5 (g cannot be 7)
==> g != f+2
==> truncate e=2,f=4; e=4,f=6; e=6,f=8
==> e+f-1 != 7
==> truncate e=2,f=6, e=6,f=2
==> e+f-1 != 9
==> truncate e=2,f=8; e=8,f=2; e=4,f=6; e=6,f=4
==> listing solution(s)
e f a g d
2 5 6 4 8
4 5 8 6 2
163
987
254
183
927
456
==> 2 solutions found under SCENARIO 2
dunno if this helps. just playing around with algebra.
[/b] -
which part u dun understand can clarify with me.Originally posted by applesweetgirl:thanks very much! although i do not really understand your scenario 2...you use maths calculator izzit? some parts i really dunno how you get... but really thanks! you are darn smart! ^^
maybe need to do hand-written then can see clearer cos presentation is not nice cos of forum formatting. lols.
no maths calculator, i jus woke up, saw ur qn, and seh seh start typing away and got the answer. this kind cannot use calculator bah... -
oh yeah, when i seh seh type, i tend to get a bit si ban and computerised cos trying to force myself to wake up..... that's how my brain works lah. so i'm like writing in programming code.
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SCENARIO 2 (b=3 and c=9)
==> 1a3
==> 9d7
==> efg
if u understood scenario 1, and basically i jus switched b=2 with b=3 and c=8 with c=9 and continued to solve.
a+d+10 = d+e+f+9 = d+f+g+7
==> g = e+2
==> a = e+f-1
all 4 2x2 squares have the same sum. try to find relations so that i can reduce the number of variables needed to solve the qn. here:
a+d+10 = d+e+f+9
==> remove d/divide by d
a+10 = e+f+9
a = e+f-1
d+e+f+9 = d+f+g+7
==> remove d/divide by d
==> remove f/divide by f
g+7 = e+9
g = e+2
==> 1 e+f-1 3
==> 9 d 7
==> e f e+2
this is a table of the simplified values. here, a and g are replaced by (e+f-1) and (e+2) respectively, as we have deduced from the previous line.
try editing this post to see the proper formatting.
==> undefined numbers 2,4,5,6,8 since we have 1,3,7 and 9 inside alrd.
==> need to define d,e,f through trial and error basically when i say d here, it's redundant. d is totally unrelated to e and f, and will be solved last. it is like a 'wildcard', being in the centre. (dun understand nvm, the word 'wildcard' is a logical outcome, not mathematical.
==> e+f-1 <= 9 logical: no digit shld exceed 9. in fact, i made a mistake. shld be 'e+f-1 <= 8' since 9 is taken. did not affect integrity of answer.
==> e+f <= 10 e+f <= 9
==> e != 5 (g cannot be 7) cos if e is 5, the position of g will become 7, but 7 is taken.
==> g != f+2 pai seh, shld be 'f != e+2' (i actually had a extra variable h just now that was an oversight, so was converting everything back.) this line is required cos if f = e+2, then f = g, then we have 2 same digits in the square.
==> truncate e=2,f=4; e=4,f=6; e=6,f=8
==> e+f-1 != 7 7 is taken.
==> truncate e=2,f=6, e=6,f=2
==> e+f-1 != 9 9 is taken
==> truncate e=2,f=8; e=8,f=2; e=4,f=6; e=6,f=4
==> listing solution(s) keeping in mind, e+f <= 9. d is listed last, and will fill in as the last digit remaining in the set of 2,4,5,6 and 8.
e f a g d
2 5 6 4 8
4 2 5 6 8
4 5 8 6 2
163 183 153
987 927 987
254 456 426
==> 3 solutions found under SCENARIO 2 -
are u still in sec skool?if u r i bet u can get A1 for A mathsOriginally posted by ^tamago^:oh yeah, when i seh seh type, i tend to get a bit si ban and computerised cos trying to force myself to wake up..... that's how my brain works lah. so i'm like writing in programming code. ...i still rmb i pri skool i luv maths...coz got problem sums 
now i sec see algebra esp in A maths i sian liao....dats y do damn badly for maths -
grad le, got A1 for both maths, but still 'o' level grad, didn't study further liao. lols.
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so smart. i can never do this in 100 years. -
man...you are really smart... i just want to clarify this part...
==> undefined numbers 3,4,5,6,9
==> need to define d,e,f through trial and error
==> e+f-1 <= 9
==> e+f <= 10
==> e != 6 (g cannot be 7)
==> f != e+1
==> truncate e=3,f=4; e=4,f=5; e=5,f=6
==> e+f-1 != 7
==> truncate e=3,f=5; e=5,f=3
==> e+f-1 != 8
==> truncate e=3,f=6; e=5,f=4
amaths??? >"< i've never done amaths before...so i find it very hard!! >""< -
die. one more solution popped out under scenario 1. interestingly, only the middle row has been changed.Originally posted by applesweetgirl:man...you are really smart... i just want to clarify this part...
==> undefined numbers 3,4,5,6,9
==> need to define d,e,f through trial and error
==> e+f-1 <= 9
==> e+f <= 10
==> e != 6 (g cannot be 7)
==> f != e+1
==> truncate e=3,f=4; e=4,f=5; e=5,f=6
==> e+f-1 != 7
==> truncate e=3,f=5; e=5,f=3
==> e+f-1 != 8
==> truncate e=3,f=6; e=5,f=4
amaths??? >"< i've never done amaths before...so i find it very hard!! >""<
too long ago liao, not seasoned yet.
SCENARIO 1 (b=2 and c=8 )
==> 1a2
==> 8d7
==> efg
good, we got the sqaure with 5 vars left.
a+d+9 = d+e+f+8 = d+f+g+7
1a2 1a2 1a2 1a2
8d7 8d7 8d7 8d7
e fg e fg efg e fg
==> g = e+1
d+e+f+8 = d+f+g+7
==> divide by d and f throughout
e+8 = g+7
g = e+1
==> a = e+f-1
a+d+9 = d+e+f+8
a+9 = e+f+8
a = e+f-1
==> 1 e+f-1 2
==> 8 d 7
==> e f e+1
new table. note that g is now (e+1) and a is now (e+f-1).
==> undefined numbers 3,4,5,6,9 cos we have 1,2,7 and 8 inside alrd.
==> need to define d,e,f through trial and error our remaining vars. d will be last cos we know least about it.
==> e+f-1 <= 9 this digit cannot exceed 9.
==> e+f <= 10 easier to read
==> e != 6 (g cannot be 7) if e is 6, g=e+1=7, but 7 is taken.
==> f != e+1 if f=e+1, then f=g, but we cannot have duplicate digits.
==> truncate e=3,f=4; e=4,f=5; e=5,f=6
==> e+f-1 != 7 digit 7 is taken. e+f != 8
==> truncate e=3,f=5; e=5,f=3
==> e+f-1 != 8 digit 8 is taken. e+f != 9
==> truncate e=3,f=6; e=6,f=3; e=5,f=4; e=4,f=5
==> listing solution(s) taking into account e+f <= 10. (3,4) out, (3,5) out, (3,6) out, (4,3) in, (4,5) out, (4,6) in, (5,3) out, (5,4) out.
e f a g d
4 6 9 5 3
4 3 6 5 9
192 162
837 897
465 435
==> 2 solutions found under SCENARIO 1 -
die. one more solution popped out under scenario 1. interestingly, only the middle row has been changed.Originally posted by applesweetgirl:man...you are really smart... i just want to clarify this part...
==> undefined numbers 3,4,5,6,9
==> need to define d,e,f through trial and error
==> e+f-1 <= 9
==> e+f <= 10
==> e != 6 (g cannot be 7)
==> f != e+1
==> truncate e=3,f=4; e=4,f=5; e=5,f=6
==> e+f-1 != 7
==> truncate e=3,f=5; e=5,f=3
==> e+f-1 != 8
==> truncate e=3,f=6; e=5,f=4
amaths??? >"< i've never done amaths before...so i find it very hard!! >""<
too long ago liao, not seasoned yet.
SCENARIO 1 (b=2 and c=8 )
==> 1a2
==> 8d7
==> efg
good, we got the sqaure with 5 vars left.
a+d+9 = d+e+f+8 = d+f+g+7
1a2 1a2 1a2 1a2
8d7 8d7 8d7 8d7
e fg e fg efg e fg
==> g = e+1
d+e+f+8 = d+f+g+7
==> divide by d and f throughout
e+8 = g+7
g = e+1
==> a = e+f-1
a+d+9 = d+e+f+8
a+9 = e+f+8
a = e+f-1
==> 1 e+f-1 2
==> 8 d 7
==> e f e+1
new table. note that g is now (e+1) and a is now (e+f-1).
==> undefined numbers 3,4,5,6,9 cos we have 1,2,7 and 8 inside alrd.
==> need to define d,e,f through trial and error our remaining vars. d will be last cos we know least about it.
==> e+f-1 <= 9 this digit cannot exceed 9.
==> e+f <= 10 easier to read
==> e != 6 (g cannot be 7) if e is 6, g=e+1=7, but 7 is taken.
==> f != e+1 if f=e+1, then f=g, but we cannot have duplicate digits.
==> truncate e=3,f=4; e=4,f=5; e=5,f=6
==> e+f-1 != 7 digit 7 is taken. e+f != 8
==> truncate e=3,f=5; e=5,f=3
==> e+f-1 != 8 digit 8 is taken. e+f != 9
==> truncate e=3,f=6; e=6,f=3; e=5,f=4; e=4,f=5
==> listing solution(s) taking into account e+f <= 10. (3,4) out, (3,5) out, (3,6) out, (4,3) in, (4,5) out, (4,6) in, (5,3) out, (5,4) out.
e f a g d
4 6 9 5 3
4 3 6 5 9
192 162
837 897
465 435
==> 2 solutions found under SCENARIO 1 -
there are now 3 solutions for scenario 2.
163 183 153
987 927 987
254 456 426
okies, the solution is completed. u can read back my earlier post again.
i was wondering if the 2 missing squares were keeping you from understanding the solution proposed.
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Sodoku different from what thread starter wants to find.Originally posted by FireIce:maybe u shd buy the sudoku game and play -
SMLJ. I cant understand a this. This question is for what lvl de? Pri, Sec, JC, Poly, Uni or what???
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wth men..
my maths ..is CMI~..worse sub..of all -
chim...really chim...got aliens among us...
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I look at maths I sian alrdy