Four Digit Numbers from 2x2 Digits - Simple Math Puzzle

[bob012345]

TL;DR

How many four digit numbers are the product of two two digit numbers?

Of all the possible four digit numbers how many are the produce of two two digit numbers? Can the process be easily generalized to other combinations? Thanks.

 

[phinds]

Summary:: How many four digit numbers are the product of two two digit numbers?

None. Think about it. (No, don't think about it. In fact, I was never here )

 

[mfb]

None. Think about it.

50*50=2500?

@bob012345: I don't know if this is homework or not, but either way you should at least make some effort to find an answer. It's really not difficult. In what range can the first factor be? For a given first factor, what range can the second factor have? Do you count different ways to get products as different results or not? E.g. 20*80 vs. 40*40.

 

[phinds]

50*50=2500?

Looks like I slipped a digit somewhere

(Actually, I know where but I'm embarrassed to say)

 

[bob012345]

50*50=2500?

@bob012345: I don't know if this is homework or not, but either way you should at least make some effort to find an answer. It's really not difficult. In what range can the first factor be? For a given first factor, what range can the second factor have? Do you count different ways to get products as different results or not? E.g. 20*80 vs. 40*40.

Thanks. That counts as one case as do permutations. I don't care how many ways get to 1600 for example as long as 1600 is only counted once. I made this up for fun and it's not homework. I have an answer but if I gave the answer upfront it wouldn't be a puzzle! Also, I want to see if anyone has a better than brute force method.

Also, for the simpler case of how many two digit numbers are the product of two one digit numbers, the answer is 32 out of a total 90 possible two digit numbers.

 

[bob012345]

View attachment 266430
Looks like I slipped a digit somewhere

(Actually, I know where but I'm embarrassed to say)

No problem.

 

[bob012345]

The answer I get is that there are 3339 numbers that are a product of two double digit numbers out of possible 9000 four digit numbers (1000 to 9999 inclusive). But there could be a counting error. I think it will scale by 100 for each higher level (xxx*xxx=xxxxxx, xxxx*xxxx=xxxxxxxx ect.) but also grow slightly because the 'grid' is finer as the ratio of allowable numbers/total possibilities converges to an asymptotic limit i.e. 32/90=0.3556, 3339/9000=0.371, ...).

 

[pbuk]

Also, for the simpler case of how many two digit numbers are the product of two one digit numbers, the answer is 32 out of a total 90 possible two digit numbers.

I don't think it is: are you counting the squares twice?

 

[bob012345]

I don't think it is: are you counting the squares twice?

That would be 58. You can easily do it graphically and see you get 32.

 

[pbuk]

58 - 32 = 26. There aren't 26 2 digit squares so I clearly don't understand the problem. What are your 32?

 

[bob012345]

58 - 32 = 26. There aren't 26 2 digit squares so I clearly don't understand the problem. What are your 32?

The bold is the diagonal elements. Originally, I was thinking no permutations which I mean upper and lower elements around the diagonal but I did forget that several elements also have duplicate values so my number was too high. It should be 27. I think the problem is better framed and my answer for the original case would be correct if I allowed multiple instances of the same value as long as it wasn't from a permutation of the same two numbers. Sorry about that!

9		18	27	36	45	54	63	72	81
8		16	24	32	40	48	56	64	72
7		14	21	28	35	42	49	56	63
6		12	18	24	30	36	42	48	54
5		10	15	20	25	30	35	40	45
4			12	16	20	24	28	32	36
3				12	15	18	21	24	27
2					10	12	14	16	18
1
	1	2	3	4	5	6	7	8	9

 

[phinds]

 

[pbuk]

You have 3 x 6 = 18 and 2 x 9 = 18.

 

[pbuk]

Spoiler

 

[bob012345]

Spoiler

Thanks! I forgot about identical values from non-permutations. Sorry.

But, I could not do the original case by hand if I didn't allow duplicates!

 

[phinds]

You have 3 x 6 = 18 and 2 x 9 = 18.

Yeah. OOPS. I misinterpreted the question

 

[pbuk]

But, I could not do the original case by hand if I didn't allow duplicates!

When posing puzzles you either need to cross-check your solutions using a different method or show them to a friend or small group first to check.

 

[phinds]

But, I could not do the original case by hand if I didn't allow duplicates!

I don't wish to be rude, but that's ridiculous. If you had done it by hand, the duplicated should have been obvious and you should have eliminated them from the total.

When I did it, I just looked to see if the boxes contained a 2-digit # and didn't even think about WHAT the number was, or duplicates, which was wrong.

 

[bob012345]

I don't wish to be rude, but that's ridiculous. If you had done it by hand, the duplicated should have been obvious and you should have eliminated them from the total.

When I did it, I just looked to see if the boxes contained a 2-digit # and didn't even think about WHAT the number was, or duplicates, which was wrong.

The original problem came first and had thousands of cases. I did not think about duplicates because I only thought about getting 4 digits but I did think about permutations. Later, when I was asked about duplicates, I should've just said they are ok. At that point I didn't consider the need to evaluate each value. So, if I had thought of that I would not have done the problem by hand (without writing script). I added the simpler case later and only then did I think about the problem of duplicates.

Sorry again about the confusion.

 

[bob012345]

When posing puzzles you either need to cross-check your solutions using a different method or show them to a friend or small group first to check.

Sorry, I should have just posed it as a problem and not a puzzle which assumes it was completely understood!

 

[phinds]

The original problem came first and had thousands of cases. I did not think about duplicates ...

A reasonable point. I was only thinking of the smaller case.