123456789 = 2002?

Given the equation

123456789 = 2002

place + or * (plus or multiply) symbols as needed between the digits on the left side of the equation such that the equation is true.

12 + 3 x 456 + 7 x 89 is nicely close but not quite there. ^_^

((1+4+6+8+9)*5+3)*7*2=2002

Originally posted by bindasraka
((1+4+6+8+9)*5+3)*7*2=2002

Re-ordering the digits is not allowed.
Using parenthesis is not allowed - only the symbols + and *.

Originally posted by SwissGambit
Re-ordering the digits is not allowed.
Using parenthesis is not allowed - only the symbols + and *.

Aintchu the party pooper🙂 You didn't spec that one in your first post!

Originally posted by SwissGambit
Given the equation

123456789 = 2002

place + or * (plus or multiply) symbols as needed between the digits on the left side of the equation such that the equation is true.

Check me if wrong:

I believe there are 2^8 possible arrangements of * and + between the 9 numerals. Call these the *+ arrangements. Example: **+*++**. That is 256 testable arrangements. Certain *+ arrangements can be excluded, such as any group of 8 that ends with ***, because 6*7*8*9>2002, no matter how 12345 are operated upon. It can also be noted that by the rules of math, multiplication supervenes upon addition, such that a*b+c*d works out as (a*b)+(c*d), not as a*(b+c)*d.

Firefox allows such numeric expressions to be put in the search bar and with an = sign, will deliver the result. It uses the above math rule about * and +.

I see no way to avoid a tedious brute force approach other than that there are some exclusions available like the above.

We are working in base 10, right?

Is there any significance to the arrangement that yields 2002? Or could any of the 256 possibilities have been presented? It's an easy puzzle to construct, if the latter.

Related question: is there only one *+ arrangement that yields 2002?

Can the general conjecture that no two *+ arrangements yield the same number, be proven or disproven, using a non-brute force method of proof?

It would be easier to work on a smaller set, like 1234. But do the results extrapolate?

I find these questions to be more interesting than the puzzle presented, but they wouldn't come up without the puzzle being presented, so thanks.

Aren't there 3^8 possibilities? Any two adjacent numbers can have a plus sign, multiplication sign, or neither between them, so it's 1 + 2, 1 x 2 or 12 ?

Originally posted by talzamir
Aren't there 3^8 possibilities? Any two adjacent numbers can have a plus sign, multiplication sign, or neither between them, so it's 1 + 2, 1 x 2 or 12 ?

It's hard to construct a puzzle that is unambiguous. I suppose it's up to SwissGambit to arbitrate, but I think when it says "between the digits" it means the LHS is to be considered as consisting of digits, not numbers, which are -- when above 9 -- combinations of digits). Of course it can be played the way you suggest, too. Similar results, I think, in terms of brute force being needed (immensely moreso!) and similar questions. What fun! (PS I was certified to teach HS math but never did aside from my internship.)

"A digit is a symbol (a numeral symbol such as "3" or "7" ) used in combinations (such as "37" ) to represent numbers in positional numeral systems." (Wikipedia)

Originally posted by SwissGambit
Re-ordering the digits is not allowed.
Using parenthesis is not allowed - only the symbols + and *.

1*23+45*6*7+89 = 2002

Originally posted by sonhouse
Aintchu the party pooper🙂 You didn't spec that one in your first post!

Would it have made you feel better if I had used the word 'only' to rule out all other means of making the equation true? 😉

Originally posted by bindasraka
1*23+45*6*7+89 = 2002

That's one. 🙂

Originally posted by JS357
Check me if wrong:

I believe there are 2^8 possible arrangements of * and + between the 9 numerals. Call these the *+ arrangements. Example: **+*++**. That is 256 testable arrangements. Certain *+ arrangements can be excluded, such as any group of 8 that ends with ***, because 6*7*8*9>2002, no matter how 12345 are operated upon. It can also be noted that b ...[text shortened]... the puzzle presented, but they wouldn't come up without the puzzle being presented, so thanks.

It's 3^8 possibilities as pointed out by talzamir.

Yes, base 10. I wouldn't post a problem in another base without specifying it.

I may comment on puzzle construction later. 🙂

There are two arrangements that yield 2002.

I don't know if a non-brute method can prove uniqueness of arrangement.

Originally posted by JS357
It's hard to construct a puzzle that is unambiguous. I suppose it's up to SwissGambit to arbitrate, but I think when it says "between the digits" it means the LHS is to be considered as consisting of digits, not numbers, which are -- when above 9 -- combinations of digits). Of course it can be played the way you suggest, too. Similar results, I think, in terms ...[text shortened]... ons (such as "37" ) to represent numbers in positional numeral systems." (Wikipedia)

talzamir is right. If no symbol is placed between digits, they read as a single number. But I'm not sure why that's ambiguous - what would you do with the numbers with no symbol between them if they are distinct numbers and not just digits? For example, what does the expression 1 2 3 = 2002 even mean?

Originally posted by SwissGambit
It's 3^8 possibilities as pointed out by talzamir.

It has to be; there are no solutions with only + and *, and no multi-digit numbers.

Yes, base 10. I wouldn't post a problem in another base without specifying it.

Although from a certain point of view, this was also a ternary problem!

I don't know if a non-brute method can prove uniqueness of arrangement.

I doubt that. Brute forcing is simple, of course. I did this one in a Spectrum emulator 🙂

Richard

Originally posted by Shallow Blue
It has to be; there are no solutions with only + and *, and no multi-digit numbers.

Yes, base 10. I wouldn't post a problem in another base without specifying it.

Although from a certain point of view, this was also a ternary problem!

I don't know if a non-brute method can prove uniqueness of arrangement.

I doubt that. Brute forcing is simple, of course. I did this one in a Spectrum emulator 🙂

Richard

What is a spectrum emulator?