help!Two questions from australian math competition(senior)

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viva0yue999
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1.For each vertex of a cube,a plane is constructed through the three vertices which are neighbours of that vertex.Into how many parts do these eight planes dissect the cube?
a)9 b)13 c)21 d)27 e)24

2. A 3*3 square is divided up into nine 1*1 unit squares. Different integers from 1 to 9 are written in these 9 unit squares. For each two squares sharing a common edge, the sum of the integers in them are calculated. The minimum possible number of different sums is
a)3 b)4 c)5 d)6 e)7

3.When the interger is divided by 23,the remainder is 11,when divided by 24,the remainder is 12 and when divided by 25,the remainder is 13.What is this integer?(I cannot remember the exact numbers in this question,but just want to know the rough idea to tackle this question).


Thank you~
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beauford
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(Original post by viva0yue999)
1.For each vertex of a cube,a plane is constructed through the three vertices which are neighbours of that vertex.Into how many parts do these eight planes dissect the cube?
a)9 b)13 c)21 d)27 e)24

2. A 3*3 square is divided up into nine 1*1 unit squares. Different integers from 1 to 9 are written in these 9 unit squares. For each two squares sharing a common edge, the sum of the integers in them are calculated. The minimum possible number of different sums is
a)3 b)4 c)5 d)6 e)7

3.When the interger is divided by 23,the remainder is 11,when divided by 24,the remainder is 12 and when divided by 25,the remainder is 13.What is this integer?(I cannot remember the exact numbers in this question,but just want to know the rough idea to tackle this question).


Thank you~
To tackel 3, note that 23,24 and 25 are coprime.
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JamesF
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For 3, add 12 to that number first of all.
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beauford
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(Original post by viva0yue999)
2. A 3*3 square is divided up into nine 1*1 unit squares. Different integers from 1 to 9 are written in these 9 unit squares. For each two squares sharing a common edge, the sum of the integers in them are calculated. The minimum possible number of different sums is
a)3 b)4 c)5 d)6 e)7


Thank you~
I believe (although I may be wrong) that this is 6. It's trivial to show that it's at least 4, i can't prove it for 6 yet.
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JamesF
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(Original post by beauford)
I believe (although I may be wrong) that this is 6. It's trivial to show that it's at least 4, i can't prove it for 6 yet.
What i'm trying now, is that if there is 6, then there are 8 sums which have only 2 different values, so 2 different numbers must be able to be made in 4 ways, and this can put a bound on the sum, because small numbers can only be made in a few different ways.
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Wagamuffin
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I think we would need the proper figures for question 3, because it states that we are looking for an integer, and with random numbers used to define criteria it probably wouldn't work out.
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beauford
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(Original post by Wagamuffin)
I think we would need the proper figures for question 3, because it states that we are looking for an integer, and with random numbers used to define criteria it probably wouldn't work out.
There are infinite solutions, you'd just need to see which one fitted the criteria
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viva0yue999
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any solutions?
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JamesF
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3. x+12 = 23*24*25*a for any integer a, the smallest solution is therefore when a = 1, so the smallest solution is 13788. There are infinite solutions of the form 13788 + 13800k for any integer k.
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