This is a great game both for elementary and middle school students. This game can be quite challenging. For younger students we suggest allowing them to work as a team (class against teacher), older students may work alone or in pairs. Who Did It? can be a great choice for a game day or a substitute. It may take up an entire class period but is well worth it! Students will be engaged, thinking, and building problem-solving and multiplication/factor skills.

Determine the number of bones based on the level of your students. The more bones, the more challenging the game. The activity leader will select and arrange their bones in secret. This number is the culprit. For added fun, provide a backstory such as this number robbed a bank or this number took a pencil from the teacher's lounge, etc. The activity leader should also pick a "crime", the crime is the row we will give clues for.

Each round, students will use clues to create a list of suspects, ultimately narrowing it down until someone determines who did "it". The clues will always be the same:

• One clue for each middle joint forming a parallelogram, the clue is the sum of the joint without carrying. For instance, if the joint sums to 15, the clue is 15. Carrying from the previous joint is also not permitted.

• One clue gives the sum of the first and last (triangular joint).

Note, clues can be given in any order. That is, the culprit may decide when to give players the information about the triangular joints.

Before play begins, announce the number of digits. The number of rounds (and clues) should be the number of digits. For instance, if the culprit is 312, we would allow for 3 rounds.

The activity leader will announce a clue each round. Clues are given as a joint appearing on the number.

For example, one clue may be "The third joint (hundreds place) of the culprit times 4 is 2." Using the bones and their understanding, students begin to build a suspect list.

A place value of 2 leaves us only with the options 2 and 0, 0 and 2, or 1 and 1. Since the clue told us we are multiplying by 4, this means 4 times the first rod must end in a 0, 1, or 2. Examining the 4th row of our rods leaves us with 3, 5, or 8. We can eliminate the 0 rod, as our first digit cannot be 0.

Narrowing down their suspect list, if their first rod is the 3, the second digit must have a 0 in the tens place of the 4 column. Leaving us with 0, 1, or 2. If their first rod is the 5, the second digit must have a 2 in the tens place, giving them 6 or 7 (as we are playing with a half-set of bones, the 5 cannot be repeated). Finally, if the first digit is 8, the second bone must have a 0 in the tens place, again leading to 0, 1, or 2.

Encourage students to share their reasoning with their team. In the case above, as the multiples of 4 that end in a 0 are 12, 20, and 32, they can conclude that for 12 and 32 (ending in a 2) must be paired with a 1-digit multiple of 4, while 20 must be paired with a value whose multiple of 4 is in the twenties to find 20, 24, 28 or 5, 6, and 7 rods.

Provide students time to create their suspect list. After the first clue, they should have narrowed things down to:

• 30_

• 31_

• 32_

• 56_

• 57_

• 80_

• 81_

• 82_

Once the round is completed, the culprit will leave another clue. As we are working with a 3-digit number, the best clue left is "the second joint (tens place) of the culprit times 4 sums to 4."

Time to play with the bones. Our second digit must be 0, 1, 2, 5, 6, or 7. We should consider the multiples of each of these values and 4 to find 0, 4, 8, 20, 24, and 28. Notice both 8 and 28 can be thrown out as they are too big (larger than 4). This means our second digit cannot be a 2 or a 7, and we eliminate these from our suspect list.

• 30_

• 31_

• 56_

• 80_

• 81_

Continuing to narrow down their list, 30 would require a final bar with a 4 in the tens place. The only such bar would be a 10-bar (or higher), which we do not have and can eliminate this option. 31 would require a 1-digit multiple of 4 for the final bar, narrowing it down to 0 or 2. 56 would also require a 1-digit multiple of 4 leaving 0, 1, or 2. 80 would require a 4 in the tens place and can thus be eliminated. Finally, 81 would require a 1-digit multiple of 4 allowing for 0 or 2.

• 310

• 312

• 560

• 561

• 562

• 810

• 812

It's time for the culprit to provide the final clue. In this game, the only clue left is the sum of the triangular numbers, which the culprit tells us is 9. There are many problem-solving methods, as there are for the previous clues. We can consider the possible ways to make 9 (1+8, 2+7, ...), we could also consider the 3-digit multiples of 4. From our suspect list, a culprit in the 3-hundreds will have a thousands place of 1, as 4×3=12, a culprit in the 5-hundreds will have a thousands place of 2, as 5×4=20, and a culprit in the 8-hundreds will have a thousands place of 3, as 8×4. That leads us to only a few options. If 310 or 312, the last digit must be an 8 (as 1+8=9). This is not the case with 310, so we can mark them from our list. If our culprit is in the 5-hundreds, the ones place must be a 7 (as 2+7=9). However, no multiple of 4 can end in a 7, which eliminates all the 500 values from our list. Finally, if our culprit is 810 or 812, the ones place must have a 6 (as 3+9=6). As 0×4=0 and 2×4=8, neither value will work. We have found the culprit! It must be 312.

There are many problem-solving approaches throughout this game. Not only are students working with the bones, considering possible values and multiples, they are also using logic skills. Students greatly enjoy a "whodunnit" so this is a fun game to get them thinking, playing with multiples and engaging in mathematics.