How it Works › Forums › Introductory Problems (AMC 8 Problems 1020) › Counting Cleverly Problem Set 1!
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Hey Math Circle Students!Here are the introductory problems for the next $3$ days! These are more difficult than AMC 8 Problems 1015 but they fit the range of $1621$. Each of these problems are worth $5$ points to your leaderboard score! If you don’t understand one of the questions, feel free to post in the questions forum here
1. How many $3$ digit numbers have $4$ as their largest digit and are divisible by both $3$ and $5$?
2. On an 8×8 chessboard that is in a checkerboard black and white pattern, how many rooks can you place on the board without any one rook attacking any other (A rook attacks squares that are either on its same row or column). Similarly, how many bishops can you place without any bishop attacking any other? (A bishop can attack any other piece that is on its same diagonal)
3. How many squares are present in a 5×5 grid of squares. This includes squares of any size: 1×1, 2×2, …etc.
Remember that you can post your solutions below to these problems and I will keep a running leaderboards for the top 10 students who participate the most! I especially appreciate the same problem being solved in different ways!
Anton1. Since it must be divisible by 5, and all of the digits are 4 or less, then the ones digit must be a 0. Next, since it must be divisible by 3, the sum of the digits must be a multiple of 3. 4 must be in the number, in either the hundreds place or tens place. To make it so that the sum of digits is a multiple of 3, the last digit must be 2 or 5, but it can’t be 5 since it is greater than 4. As a result, there are 2 numbers that satisfy this condition: 240, 420
2.There can only be 8 rooks, since each rook attacks one row and one column. We can place one bishop on one edge, since that limits the pieces the most, and they take up the least space possible. By doing this, we can fit 6 more bishops on the bottom edge, for a total of 14 bishops.
3. We can do casework:
1×1: 25
2×2: 16, We look at which spots the top left corner can be in the square
3×3: 9
4×4: 4
5×5: 1
The sum of these is 55
Tejas1) We can do what we did for problem 6 on the 10/18/16 meeting. We know the answers are 420 and 420.
2)We can fit 8 rooks, as they would have to be diagonal because any on the same line would be attacking reach other, and it is an 8×8 so the answer is 8 rooks.Also, we can have 8 bishops on the bottom of the board, and 6 on the top leads to a total of 14. We have 6 not 8 because the spaces on the ends can’t be there or else they will attack each other.
3)We can add up the number of squares by going 1×1, 2×2, and so on to 5×5. The total is 25+16+9+4+1 which is equal is to 55.Tejas and Anton, great job on problems 1 and 3! You have both received 10 points for those, but you have a slight error in problem 2. Here is what you need to work on:
1. Although you have identified a possible maximum configuration, you never showed that that was the best one.
2. How do you know that any other configuration doesn’t allow more bishops?Think about counting this in a clever way that is guaranteed to give you the maximum. But great job nonetheless!
PATYUSH MAKKAR1.(did we have to have 4 as a digit)
sol (not including 4 in some cases)BY the first constraint we have 0,1,2,3,4(remember that the last digit has to be 0)
case 1 – first digit 1 w get the sum of the digits as 1 now the 2nd digit has to be 2
case 2 2 – has to be 1,4
case 33has to be 0 or 3
case 44 – has to be 2
adding all the cases we get 6!2. we use the multiplication principle of counting
let the rook be in a random place , since the second rook cannot be in the same column or row, we have 49 choices left for the second one thus there are 64 choices for the first one, 49 for the second one. Repeating the process until we have 0 places left we get 8 rooks.3. by doing casework and looking at possible transformations we get the total as 55
AyushGreat job Pratyush! You get 10 points. But make sure to address the second part of problem 2 that is slightly more difficult.

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