Recover missing digits in a base-12 (duodecimal) number system

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jjfro9

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I am somewhat lost as where to start this problem. I think I have to use modulus 7, but not exactly sure how I would go about it. Any help would be appreciated.

Find all the missing digits represented by _ (Assume they use A for ten, B for eleven, x for multiplication, and = for equal)

209 x _ _ _ _ = 7B _ A _ 0
 
I assume the 7 is the first digit of the product, not a multiplier, so I don't see where mod 7 would come in.

You might start by thinking how you can get the 0, and the 7, in the product, which will give only a few options for the first and last digits of the second factor.
 
Thanks for the hint. That makes sense. That helped me figure out that the last number must be 0 in the second factor. I believe I have it figured out, but any hints on I would show how to figure the rest of the problem other than by brute force. The answer I came up with is 209 x 3A60 = 7BAA60
 
jjfro9 said:
Thanks for the hint. That makes sense. That helped me figure out that the last number must be 0 in the second factor. I believe I have it figured out, but any hints on I would show how to figure the rest of the problem other than by brute force. The answer I came up with is 209 x 3A60 = 7BAA60
Click to expand...

Well, the last digit doesn't necessarily have to be 0. Just as you can get a 0 in base ten not only by multiplying by 0, but also by multiplying 5*2, in base twelve you can get a 0 in a couple other ways.

Have you done the multiplication to check your answer? I don't think it works.

One thing I would do (I haven't tried very hard yet) is to make a duodecimal multiplication table to look at while working on the problem! Generally, intelligent "brute force" is the way to go -- you have to work it out bit by bit, and often check out several options, but you won't be checking every possible combination, which would be true "brute force"!
 
What do 2*6, 3*4, 3*8, 4*6, 4*9, 6*6, 6*8, 6*10 and 8*9 all have in common?
 
They are all factors of 12, or multiples of 12, which would result in a 0 in my last digit in base 12. So my last digit in _ _ _ _ could be either 0, 4 or 8 since those are my only digits that would result in a 0 when 9 is multiplied by it.?
 
Dr.Peterson said:
Well, the last digit doesn't necessarily have to be 0. Just as you can get a 0 in base ten not only by multiplying by 0, but also by multiplying 5*2, in base twelve you can get a 0 in a couple other ways.

Have you done the multiplication to check your answer? I don't think it works.

One thing I would do (I haven't tried very hard yet) is to make a duodecimal multiplication table to look at while working on the problem! Generally, intelligent "brute force" is the way to go -- you have to work it out bit by bit, and often check out several options, but you won't be checking every possible combination, which would be true "brute force"!
Click to expand...
I am still getting 209 x 3A60 = 7BAA60 as working.
I also can't find any other solutions that work because I believe I am confined to having 3 as my first digit and either 0, 4, or 8 as my last digit based on the fact that I have to multiply that 4th digit by number 209.
 
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jjfro9 said:
I am still getting 209 x 3A60 = 7BAA60 as working.
I also can't find any other solutions that work because I believe I am confined to having 3 as my first digit and either 0, 4, or 8 as my last digit based on the fact that I have to multiply that 4th digit by number 209.
Click to expand...
I had converted your three numbers to base ten using Wolfram Alpha and they didn't work out; just now I did your multiplication by hand and got 801A60. I could still be wrong, of course.

But you're right about 0, 4, 8. And I haven't had time to try actually solving the puzzle myself.
 
[MATH]209_{12} = 2 * 12^2 + 0 * 12 + 9 =\\ 2 * 144 + 0 * 12 + 9 * 1 = 297.[/MATH][MATH]3A60_{12} = 3 * 12^3 + 10 * 12^2 + 6 * 12 + 0 =\\ 3 * 1728 + 10 * 144 + 72 = 6696.[/MATH][MATH]297 * 6696 = 1988712 = 7 * 248832 + 246888 =\\ 7 * 12^5 + 11 * 20736 + 18792 = 7 *12^5 + 11 * 12^4 + 10 * 1728 + 1512= \\ 7 * 12^5 + 11 * 12^4 + 10 * 12^3 + 1440 + 72 = \\ 7 * 12^5 + 11 * 12^4 + 10 * 12^3 + 10 * 12^2 + 6 * 12^1 + 0 * 1 = \\ 7BAA60_{12}.[/MATH]Now I never guarantee that my arithmetic is correct; I have always thought arithmetic was mental drudgery of the worst kind. But at least I have put my workings down so any error can be identified. I think the OP is correct.

Of course, I did not go back to the original puzzle.
 
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jjfro9 said:
They are all factors of 12, or multiples of 12, which would result in a 0 in my last digit in base 12. So my last digit in _ _ _ _ could be either 0, 4 or 8 since those are my only digits that would result in a 0 when 9 is multiplied by it.?
Click to expand...
Factors of 12 are 1, 2, 3, 4, 6 and 12. Multiples of 12 are 0*12, 1*12, 2*12, 3*12, ... . Factors of 12 and multiples of 12 are not the same thing. My list contained multiples of 12. Of course I (accidentally) left out 0.
 
Yes, you are correct. I did type the wrong word there. Using multiples is more clear. Thanks everyone. I think I have it now. I appreciate everyone's input.
 
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