Dan Finkel
@mathforlove.bsky.social
Founder, Math for Love, mathforlove.com.
Dedicated to giving everyone a chance and a reason to fall in love with mathematics.
Curriculum writer, game developer, speaker, teacher of teachers.
Dedicated to giving everyone a chance and a reason to fall in love with mathematics.
Curriculum writer, game developer, speaker, teacher of teachers.
Proofing #SubtractionbyHeart! The new cards are looking lovely :-)
We should have these out by early 2026!
#iteachmath #mtbos
We should have these out by early 2026!
#iteachmath #mtbos
September 24, 2025 at 10:01 PM
Proofing #SubtractionbyHeart! The new cards are looking lovely :-)
We should have these out by early 2026!
#iteachmath #mtbos
We should have these out by early 2026!
#iteachmath #mtbos
I either need ones on the first die to make 2, or 2s on the second die. In the first case, that's going to force die 2 to consider of odd numbers, and die 1 to contain three blank sides and three ones.
Forgot to check the total. But I'm okay - it's 21 dots totay!
Forgot to check the total. But I'm okay - it's 21 dots totay!
September 17, 2025 at 6:32 AM
I either need ones on the first die to make 2, or 2s on the second die. In the first case, that's going to force die 2 to consider of odd numbers, and die 1 to contain three blank sides and three ones.
Forgot to check the total. But I'm okay - it's 21 dots totay!
Forgot to check the total. But I'm okay - it's 21 dots totay!
There are two other possibilities: a 2 by 3 and 3 by 2 array of ones.
Each of these is going to lead to potentially more choices. Though once one die is complete, it forces the numbers on the other die.
Let's see - let's go with the first image and see the options.
Each of these is going to lead to potentially more choices. Though once one die is complete, it forces the numbers on the other die.
Let's see - let's go with the first image and see the options.
September 17, 2025 at 6:28 AM
There are two other possibilities: a 2 by 3 and 3 by 2 array of ones.
Each of these is going to lead to potentially more choices. Though once one die is complete, it forces the numbers on the other die.
Let's see - let's go with the first image and see the options.
Each of these is going to lead to potentially more choices. Though once one die is complete, it forces the numbers on the other die.
Let's see - let's go with the first image and see the options.
Yep, the rest of this is forced.
Another solution! Still, there's more here.
Another solution! Still, there's more here.
September 17, 2025 at 6:24 AM
Yep, the rest of this is forced.
Another solution! Still, there's more here.
Another solution! Still, there's more here.
Interestingly, I could also have the second die consist entirely of ones, and get a 6 by 1 array instead.
Wait a minute... I see where this is going. This is going to force Die 1 to lose one face from every side. I'll put the 1 in on die 1, and I'll get all six 2s. And so on.
Wait a minute... I see where this is going. This is going to force Die 1 to lose one face from every side. I'll put the 1 in on die 1, and I'll get all six 2s. And so on.
September 17, 2025 at 6:22 AM
Interestingly, I could also have the second die consist entirely of ones, and get a 6 by 1 array instead.
Wait a minute... I see where this is going. This is going to force Die 1 to lose one face from every side. I'll put the 1 in on die 1, and I'll get all six 2s. And so on.
Wait a minute... I see where this is going. This is going to force Die 1 to lose one face from every side. I'll put the 1 in on die 1, and I'll get all six 2s. And so on.
Interestingly, those six spots are going to need to be arranged in a rectangular shape, since they're determined by the column and row numbers.
So I could have a 1 by 6 rectangle of ones, which could only happen if the first die was blank. That'll be my trivial solution again.
So I could have a 1 by 6 rectangle of ones, which could only happen if the first die was blank. That'll be my trivial solution again.
September 17, 2025 at 6:20 AM
Interestingly, those six spots are going to need to be arranged in a rectangular shape, since they're determined by the column and row numbers.
So I could have a 1 by 6 rectangle of ones, which could only happen if the first die was blank. That'll be my trivial solution again.
So I could have a 1 by 6 rectangle of ones, which could only happen if the first die was blank. That'll be my trivial solution again.
I forgot to fill in what the total roll is. 0 + 1 = 1. I'll fill that in.
Now there are 36 possible rolls (corresponding to the 36 empty squares in the table). I need 6 of them to be ones, 6 to be twos, and so on, up to 6 sixes.
So what are the options for having 6 ones?
Now there are 36 possible rolls (corresponding to the 36 empty squares in the table). I need 6 of them to be ones, 6 to be twos, and so on, up to 6 sixes.
So what are the options for having 6 ones?
September 17, 2025 at 6:16 AM
I forgot to fill in what the total roll is. 0 + 1 = 1. I'll fill that in.
Now there are 36 possible rolls (corresponding to the 36 empty squares in the table). I need 6 of them to be ones, 6 to be twos, and so on, up to 6 sixes.
So what are the options for having 6 ones?
Now there are 36 possible rolls (corresponding to the 36 empty squares in the table). I need 6 of them to be ones, 6 to be twos, and so on, up to 6 sixes.
So what are the options for having 6 ones?
Let's think about this. To get a one, I need a 1 on one die, and a 0 on the other. (No half dots.)
Also, I can't have a 0 on both dice, because then 0 could be an outcome. So without loss of generality, I can assume my table looks like this.
Also, I can't have a 0 on both dice, because then 0 could be an outcome. So without loss of generality, I can assume my table looks like this.
September 17, 2025 at 6:13 AM
Let's think about this. To get a one, I need a 1 on one die, and a 0 on the other. (No half dots.)
Also, I can't have a 0 on both dice, because then 0 could be an outcome. So without loss of generality, I can assume my table looks like this.
Also, I can't have a 0 on both dice, because then 0 could be an outcome. So without loss of generality, I can assume my table looks like this.
Normally I'd use pen and paper, but I think it'll be easier to take screenshots of a computer visualizations, so I'll do that for the purposes of speed in posting. Unless I get stuck, in which case I'll need to move to pen and paper to let my brain roam a little more.
September 17, 2025 at 6:10 AM
Normally I'd use pen and paper, but I think it'll be easier to take screenshots of a computer visualizations, so I'll do that for the purposes of speed in posting. Unless I get stuck, in which case I'll need to move to pen and paper to let my brain roam a little more.
This image, featuring the purple concave hexagons, blue rhombuses, and gray darts, is one of my favorites.
March 12, 2025 at 5:27 PM
This image, featuring the purple concave hexagons, blue rhombuses, and gray darts, is one of my favorites.
The 21st Century Pattern Blocks allow for even more opportunities for these kinds of provocations! We were just playing around with them in a teacher session recently.
Having blocks that can't be made from the base green triangle makes this kind of argument really appealing.
Having blocks that can't be made from the base green triangle makes this kind of argument really appealing.
March 12, 2025 at 5:26 PM
The 21st Century Pattern Blocks allow for even more opportunities for these kinds of provocations! We were just playing around with them in a teacher session recently.
Having blocks that can't be made from the base green triangle makes this kind of argument really appealing.
Having blocks that can't be made from the base green triangle makes this kind of argument really appealing.
Just something to consider.
February 6, 2025 at 11:01 PM
Just something to consider.
February 6, 2025 at 12:51 AM