Multiple of 12

Hypatia001

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Hello...

Here is the question....

how many positive numbers less than 1000000 (and greater than 1) are multiples of 12 and have only 1 or 2 or both as their digits.

I can write them all down and come up with an answer but is there an simpler mathematical way ?
 

Dr.Peterson

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how many positive numbers less than 1000000 (and greater than 1) are multiples of 12 and have only 1 or 2 or both as their digits.

I can write them all down and come up with an answer but is there an simpler mathematical way ?
I would use divisibility tests. Divisibility by 3 tells you how many of each digit there can be, and divisibility by 4 tells you what has to be in the last two places. This comes close to listing all such numbers (and may even be what you are doing), but is at least systematic, and permits counting without actually listing.
 

Hypatia001

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I would use divisibility tests. Divisibility by 3 tells you how many of each digit there can be, and divisibility by 4 tells you what has to be in the last two places. This comes close to listing all such numbers (and may even be what you are doing), but is at least systematic, and permits counting without actually listing.
I understand that and have used that in a way by listing all the possible numbers and then dividing all of those by 12. This works but i am thinking that there must be a more efficient way.

Ps... the question is being asked by my 10 year old and i have no way of explaining how this is done other than listing every single number.
 

Dr.Peterson

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I understand that and have used that in a way by listing all the possible numbers and then dividing all of those by 12. This works but i am thinking that there must be a more efficient way.

Ps... the question is being asked by my 10 year old and i have no way of explaining how this is done other than listing every single number.
In order to tell you whether there is a more efficient way to do it, I'd have to see the details of your work. The only improvements in efficiency I can imagine are in those details, not in doing it an entirely different way. There is no formula for it, unless I'm missing some obscure trick.

And for a 10-year-old, I'm even more sure. Listing is fully appropriate.
 

Hypatia001

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In order to tell you whether there is a more efficient way to do it, I'd have to see the details of your work. The only improvements in efficiency I can imagine are in those details, not in doing it an entirely different way. There is no formula for it, unless I'm missing some obscure trick.

And for a 10-year-old, I'm even more sure. Listing is fully appropriate.
Below is my scribblings. We did this together. Her brain was able to make more sense of the question than my own !!
 

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Dr.Peterson

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Below is my scribblings. We did this together. Her brain was able to make more sense of the question than my own !!
Thanks. Looks like a nice orderly list, from which you eliminated some (many) by using divisibility tests, I presume. What I would do is considerably more efficient, even if I try to keep it at the appropriate level. The key is to only imagine writing out the list, until there's a reason to do so. That is, cut down the size of the list before actually writing it.

First, since the number has to be divisible by 4, the last two digits must form a multiple of 4 (since the rest of the digits form a multiple of 100). So the last two digits have to be 12. That cuts your list down by a factor of 4.

Second, since the number has to be divisible by 3, the sum of the digits must be a multiple of 3. We can list all possibilities, ignoring order:

12, 111, 222, 1122, 11112, 12222, 111111, 111222, 222222​

(There are several ways this list can be made, at varying levels of sophistication.) Now just look for all ways to order those, putting 12 at the end.

If I were working with a child on this, I would probably watch them make a list and then point out things like "none of those numbers were even, so they couldn't work. Why don't we skip writing odd numbers from now on?" or "None of the numbers in this set of four worked; why not? Could we decide ahead of time which groups not to bother with?"

This is exactly the sort of thinking I often do while I'm working on an unfamiliar problem, just trying brute force while trying to learn things about the problem that can help me save time.
 

Hypatia001

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Thanks. Looks like a nice orderly list, from which you eliminated some (many) by using divisibility tests, I presume. What I would do is considerably more efficient, even if I try to keep it at the appropriate level. The key is to only imagine writing out the list, until there's a reason to do so. That is, cut down the size of the list before actually writing it.

First, since the number has to be divisible by 4, the last two digits must form a multiple of 4 (since the rest of the digits form a multiple of 100). So the last two digits have to be 12. That cuts your list down by a factor of 4.

Second, since the number has to be divisible by 3, the sum of the digits must be a multiple of 3. We can list all possibilities, ignoring order:

12, 111, 222, 1122, 11112, 12222, 111111, 111222, 222222​

(There are several ways this list can be made, at varying levels of sophistication.) Now just look for all ways to order those, putting 12 at the end.

If I were working with a child on this, I would probably watch them make a list and then point out things like "none of those numbers were even, so they couldn't work. Why don't we skip writing odd numbers from now on?" or "None of the numbers in this set of four worked; why not? Could we decide ahead of time which groups not to bother with?"

This is exactly the sort of thinking I often do while I'm working on an unfamiliar problem, just trying brute force while trying to learn things about the problem that can help me save time.
That makes amazing sense.... even to me!!

Let me show her in the morning and see how she gets on !!
 

Hypatia001

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Dr.Peterson

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Can i ask a simple question.... why divisible by 4 and why divisible by 3 ?

That is the question i was asked this morning and i had no answer... except the prof said so !!
I have been expecting more questions, as I presumed that you would know about the divisibility tests, and then you didn't explicitly mention them.

If a number is divisible by a product ab, then it is necessarily divisible also by a and b individually. In this case, a number divisible by 12 must also be divisible by 3 and by 4, since 12 = 3*4.

In addition, if the numbers a and b are relatively prime, meaning that they have no common factors except 1, then any number divisible by a and b individually must be divisible by their product, ab. This is true for 3 and 4; it would not be true for 2 and 6 (where, for example, 6 itself is divisible by 2 and by 6, but not by 12).

So the test for divisibility by 12 is to see whether a number is divisible by 3 and by 4. Since you haven't asked how I tested those, I will assume that you do know the tests I described. Here is a page about divisibility tests that includes all three of these (3, 4, 12): https://www.mathsisfun.com/divisibility-rules.html
 
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