Grades 9-10 Video Solutions 2023-2023_9-10

Video Transcription

Elizabeth wants to write the integers 1 to 9 in the 9 boxes shown so that the integers in any 3 adjacent boxes add up to a multiple of 3. In how many ways can she do this?  So we want to organize our numbers nicely into groups of 3 so that each group of 3 adds up to a multiple of 3.  So let's let a, b, and c represent the digits that are 0 mod 3, 1 mod 3, and 2 mod 3. That is, a represents 3, 6, or 9, b represents 1, 4, or 7, and c represents 2, 5, or 8.  Then we want a sequence in the form a, b, c, a, b, c, a, b, c in order to ensure that any 3 adjacent boxes adds up to a multiple of 3.  In order to see why, note that a, b, c will be 0 mod 3, 1 mod 3, 2 mod 3, which will sum to 0 mod 3.  b, c, a is again the same thing, c, a, b is the same thing, a, b, c is the same thing, and we continue all the way down.  If we did something like a, a, a, b, b, b, c, c, c, then the first 3, a, a, a, would satisfy our property,  b, b, b would satisfy our property, and c, c, c would satisfy our property.  But when we have a, a, a, b, starting from a, a and going to b, a, a, b would give 1 mod 3, and that does not satisfy our property.  Therefore, we can only find sequences in this form a, b, c, a, b, c, a, b, c. However, it doesn't need to be exactly a, b, c repeated.  It can also be something like a, c, b, b, c, a, or something else.  And note that there are 3 factorial, which is 6 ways, to order the letters a, b, c in some order and then repeat that.  Then, for the mod 0 numbers, 3, 6, and 9, there are 3 factorial ways to decide which order 3, 6, and 9 come in in our sequence, and the same will be true for mod 1 and mod 2.  Because in this sequence a, b, c, a, b, c, a, b, c, for example, we could either have 3, 6, 9, 3, 9, 6, 6, 3, 9, 6, 9, 3, 9, 3, 6, 9, 6, 3.  We have 6 ways, and then we can also do that with b and c. Then, we can make 6 choices 4 times.  One time in doing the order of a, b, c, one time for the mod 0 numbers, one time for mod 1, and one time for mod 2.  And because we make 6 choices 4 times, we have a, which is 6 to the 4, total ways to write numbers in the boxes.

Video Summary

Elizabeth can arrange the integers 1 to 9 in 9 boxes such that any 3 adjacent boxes sum to a multiple of 3 in \(6^4\) ways. To achieve this, she organizes the numbers into sequences of groups like \(a, b, c, a, b, c, a, b, c\), where \(a, b, c\) represent numbers equivalent to 0, 1, and 2 mod 3 respectively. There are 6 permutations of arranging \(a, b, c\), and for each residue class (numbers 3, 6, 9 for \(a\); 1, 4, 7 for \(b\); 2, 5, 8 for \(c\)), there are 6 possible arrangements. Therefore, the total arrangements are \(6 \times 6 \times 6 \times 6 = 1296\) ways.

Keywords

integer arrangement

adjacent boxes

multiple of 3

residue classes

permutations