Grades 11-12 Video Solutions 2023-2023_11-12

Video Transcription

Problem number 9. Each of the integers from 1 to 9 is to be placed in one of the 9 boxes in the  picture so that any three numbers in consecutive boxes add to a multiple of 3. Numbers 7 and 9  have already been placed and how many different ways can the remaining boxes be filled? Alright,  so for this problem we're gonna use something called modular arithmetic. For the integers 1  to 9 there are three integers equivalent to 0 mod 3 that would be 0, 3 and 6. There are three  integers equivalent to 1 mod 3 which would be 1, 4 and 7 and three integers equivalent to 2 mod 3.  So for this table to work we must have 0, 1 and 2 mod 3 consecutively in any order throughout the  boxes. You might be saying well why can't have 1, 1, 1? Well if you have 1, 1, 1 what number  is going to go next to that 1 right and there's no number that can fit in the 1 right next to  that because we have nine boxes that we need to fill out. So simplifying the table mod 3 with the  two values we already have we should have 0, 1 because 7 is equal to 1 mod 3 and then so the  third number should be 2 and then 9 is equal to 0 mod 3. So notice that 2 is in between the 1 and  the 0 because that's the only number that's gonna get us to 0 mod 3 and now the rest of the table  fills itself out. Just look at the previous two values in the table to determine the next value  so that the sum of any three consecutive terms is 0 mod 3. So now we've done that we know that  the mods are fixed and we just have to determine how many different ways are there to place the  different numbers. So there's two factorial or two ways to place the 0 mod 3 integers because we've  already placed one of them so we only have two remaining to be placed. Two factorial ways to  place the 1 mod 3 integers since we've already placed a 7 but for the 2 mod 3 there's three  integers that we still need to place and those integers are going to be right here, right here,  and this last part right here. And so in total there's 3 factorial equals 6 ways to place 2 mod  3 integers. Now using the fundamental principle of counting we do 2 times 2 times 6 to get a total of 24 ways and so the correct answer is choice E.

Video Summary

The problem involves placing integers 1 to 9 in boxes such that any three consecutive numbers sum to a multiple of 3. Utilizing modular arithmetic, numbers must satisfy 0, 1, and 2 mod 3 in sequence. Numbers 7 (1 mod 3) and 9 (0 mod 3) are pre-placed, determining the necessity for 2 mod 3 to follow. With fixed mods, there are 2 factorial ways for 0 mod 3 integers, 2 factorial for 1 mod 3, and 6 ways (3 factorial) for 2 mod 3 integers. By counting principle, the total arrangements are 24 ways to fill the boxes correctly. The answer is 24.

Keywords

integer placement

modular arithmetic

consecutive sum

arrangement counting

mathematical problem