patrick.net

A = 6666, B,C & D =0

6319
 319
  19
   9

There are two possibilities:

for a,b,c,d in itertools.product(range(0,9+1),repeat=4):
    if a1000+b100+c10+d + b100+c10+d + c10+d + d == 6666:
         print(a,b,c,d,"",a+b+c+d)

5 8 1 9 23
6 3 1 9 19


Hey, is there a way that I can paste code into Patrick.net without having to deal with the multiplication character? Also, I had to add "nbsp"s to get the indenting to work. I put the above post in "code" blocks, but that only gave me the monospaced font.

Patrick says

6319
 319
  19
   9

My answer is 6319....A=6000, B= 300, C=10, D=9...6000+300+10+9= 6319

SunnyvaleCA says

Hey, is there a way that I can paste code into Patrick.net without having to deal with the multiplication character? Also, I had to add "nbsp"s to get the indenting to work. I put the above post in "code" blocks, but that only gave me the monospaced font.

@SunnyvaleCA

If you use the "pre" tag instead of "code", then indenting should work with normal space characters instead of "nbsp".

Anything between two star characters will be surrounded by the b tag. This is convenient most of the time, but not when pasting code. What's a good solution to that?

I suppose what I should do is not change any input which is within pre or code tags, but this failed:

SunnyvaleCA says

There are two possibilities:

for a,b,c,d in itertools.product(range(0,9+1),repeat=4):
    if a1000+b100+c10+d + b100+c10+d + c10+d + d == 6666:
         print(a,b,c,d,"",a+b+c+d)

5 8 1 9 23
6 3 1 9 19


AMAZING WORK...it could be 6319 or 5819

1337irr says

A = 6666, B,C & D =0

Each letter is supposed to be a single digit.

Patrick says

6319
 319
  19
   9

Show your work (I am that nun rapping your knuckles with the ruler).

SunnyvaleCA says

There are two possibilities:

for a,b,c,d in itertools.product(range(0,9+1),repeat=4):
    if a1000+b100+c10+d + b100+c10+d + c10+d + d == 6666:
         print(a,b,c,d,"",a+b+c+d)

5 8 1 9 23
6 3 1 9 19


Ah, the iterative approach. I guess you coder geniuses have used the computer's ability do hundreds of guesses to displace the need for a math guy to derive a formula.

FortwayeAsFuckJoeBiden says

Don’t forget “10 for the big guy”

It's one equation with three unknowns. By the most elementary principles of linear algebra, this is a singular matrix, which necessarily has no unique solution.

EDIT: Every person should know mathematics including matrix algebra up to ordinary differential equations. Those who can't should perhaps be dealt something described in Phil Dick's short story "The Pre-Persons".

The Founding Fathers had it right - those who can't add, let alone figure our life, shouldn't be able to vote.

komputodo says

this puzzle was on facebook...about 1300 people gave the answer and i think they were all wrong. They do not tell you the correct answer so I decided to see what you guys say. Nobody gave the same answer that i did.

D=4 or 9
C=1 if D=9, C=5 if D=4
B=3, no solution if C=5
A=6

6319
0319
0019
0009
-------
6666

A+B+C+D = 6+3+1+9 = 19

How did I do?

Edit: SunnyvaleCA has the correct answer. My error was I assumed there was only a single solution, but I think using a computer is cheating. It gives you an excuse not to think.

SunnyvaleCA says

There are two possibilities:

for a,b,c,d in itertools.product(range(0,9+1),repeat=4):
    if a1000+b100+c10+d + b100+c10+d + c10+d + d == 6666:
         print(a,b,c,d,"",a+b+c+d)

5 8 1 9 23
6 3 1 9 19

That's cheating! You're supposed to do it in your head!

But that foreign language does look pretty cool.

Patrick says

6319
319
19
9

That works. Brute force?

I started by trying to get my right column remainders to be six and figure out how to make my carry' s plus "some number times three" to equal a remainder of six with some carry to make the next column...

That's when I gave up and scrolled down..

Onvacation says

That works. Brute force?

I did it in my head with the same reasoning richwicks describes above. I didn't look for multiple solutions though. Just figured that D had to be 9, and it flowed from there.

My mother was way better than I am at arithmetic. She would add up the total at the grocery store in her head before the cashier did it and be pretty much correct, often exactly correct.

Did I misunderstand the question? A+B+C+D is either 6319 or 5819 in my mind..not 19, not 23...convince me that I am wrong. Also I didn't understand what Patricks answer was all about...6319, 319, 19, 9. They didn't ask what is B+C+D or what is C+D or what is D. They just asked what is A+B+C+D

Lol, great point!

I found a solution for A, B, C, and D but I never answered the actual question, which asks what is A + B + C + D!

So I say the answer to that is 19.

HeadSet says

Ah, the iterative approach. I guess you coder geniuses have used the computer's ability do hundreds of guesses to displace the need for a math guy to derive a formula.

I was wondering about a formula. One difficulty is that you have one giant equation for all 4 variables followed by inequalities on the 4 variables separately. (inequalities such as a >= 1, a <=9, b >= 1, etc). I'm really not sure how to go about solving this mathematically.

There is nothing that states that A, B C and D need to be integer values. It's implied (and a safe assumption) but not clearly stated.

Thus

A = 6 / 1 = 6
B = 6 / 2 = 3
C = 6 /3 = 2
D = 6/4 = 1.5

6 + 3 + 2 + 1.5 = 12.5

As a check:
6321.5
+ 321.5
+ 21.5
+ 1.5
= 6666.0

komputodo says

Did I misunderstand the question? A+B+C+D is either 6319 or 5819 in my mind..not 19, not 23...convince me that I am wrong. Also I didn't understand what Patricks answer was all about...6319, 319, 19, 9. They didn't ask what is B+C+D or what is C+D or what is D. They just asked what is A+B+C+D

You must have misunderstood the question. It's asking:

ABCD +
 BCD +
  CD +
   D
----
6666

What A, B, C, and D are is unknown, but each digit would have to be between 0 and 9

Once you figure out what A, B, C, and D are, what is A+B+C+D ?

SunnyvaleCA came up with the TWO possible solutions here. I found the first solution, but not the second, so it's sour grapes for me. They are

A=6 B=3 C=1 D=9; which when added together gives you 19
A=5 B=8 C=1 D=9; which when added together gives you 23

So 23 and 19 are the answers. I don't think the person that made up the question realized there were two possible solutions.

Funny I still sometimes act like a student, and expect a single answer. It didn't even occur to me there might be two, but in a real world situation, I would assume there was 0, 1 or more.

richwicks says

What A, B, C, and D are is unknown, but each digit would have to be between 0 and 9

The digit for A is written as a 6 but it represents 6000 in the equation is what I'm saying. Therefore in the equation A=6000 not 6. When people say that a person earns 6 figures, don't they mean that they earn 100,000 or more? They don't mean 1 or more (1+0+0+0+0+0)

SunnyvaleCA says


for a,b,c,d in itertools.product(range(0,9+1),repeat=4):
    if a1000+b100+c10+d + b100+c10+d + c10+d + d == 6666:
         print(a,b,c,d,"",a+b+c+d)


@SunnyvaleCA - that Python?

komputodo says

The digit for A is written as a 6 but it represents 6000 in the equation is what I'm saying.

A in the first equation only represents a single digit, what follows it is what 10's column it is in.

At first, I thought it just wanted to know what digits A,B,C and D had to be. The question of "what is A+B+C+D" was superfluous in my opinion as that's a trivial calculation and just adds ambiguity to the problem.

Without the second part to the question, I would have said A=6000, not 6 so I understand your confusion.

Hmmmm...

5000 + 800 + 10 + 9 = 5819
6000 + 300 + 10 + 9 = 6319

Might be correct answers. I don't think the person that asked it did too much thinking about it.

I mean REALLY "ABCD" means A TIMES B times C times D with standard math notation. Now I wonder if that's what they meant. If it was in lower case and I was in Linear Algebra class, that is what I would assume they meant.

That has no solution:

#!/usr/bin/perl -w

exit (main (@ARGV));

sub main
{
  for my $a (0..9)
  {
    for my $b (0..9)
    {
      for my $c (0..9)
      {
        for my $d (0..9)
        {
          if ((($a*$b*$c*$d) + ($b*$c*$d) + ($c*$d) + $d) == 6666)
          {
            printf ("%d %d %d %d\n", $a, $b, $c, $d);
          }
        }
      }
    }
  }
  
  return 0;
}

Patrick says

Lol, great point!

I found a solution for A, B, C, and D but I never answered the actual question, which asks what is A + B + C + D!

So I say the answer to that is 19.

how can it be 19? By that reasoning, a+b+c+d+b+c+d+c+d+d would equal 6666 but it doesn't...It equals 51. Am I fucked up? In elementary school we were taught that the 1st column is 1s, the second is 10's the 3rd is 100's and the 4th is 1000's

komputodo says

how can it be 19? By that reasoning, a+b+c+d+b+c+d+c+d+d would equal 6666 but it doesn't...It equals 51.

Would you prefer the question to be:


  A * 1000 + B * 100 + C * 10 + D
           + B * 100 + C * 10 + D
                     + C * 10 + D
                              + D
---------------------------------
  6 * 1000 + 6 * 100 + 6 * 10 + 6 = 6666


Then: "what is A+B+C+D?"

That's the real question. I think. It is poorly worded, and it has two solutions, and if I used normal mathematical notation, that's not the question, and if I use normal mathematical notation, there is no solution, at least for integers and if I don't use integers, there's an infinite number of solutions.

richwicks says

komputodo says

how can it be 19? By that reasoning, a+b+c+d+b+c+d+c+d+d would equal 6666 but it doesn't...It equals 51.

Would you prefer the question to be:


  A 1000 + B 100 + C 10 + D
+ B 100 + C 10 + D
+ C 10 + D
+ D
-------------------------------
6 1000 + 6 100 + 6 * 10 + 6


Then: "what is A+B+C+D?"

That's the real question. I think. It is poorly worded, and it has two solutions, and if I used normal mathematical notation, that's not the question, and if I use normal mathematical notation, there is no solution, at least for integers and if I don't use integers, there's an infinite number of solutions.

I think the question is just fine how it is written. It just seems to me that people are trying to overcomplicate it and that they have forgotten elementary school math.

komputodo says

I think the question is just fine how it is written. It just seems to me that people are trying to overcomplicate it and that they have forgotten elementary school math.

Show your work.

Maybe I am entirely misunderstanding the question.

Oh - I think I see the problem - where do you come up with this equation:

a+b+c+d+b+c+d+c+d+d

? I think that's where the discrepancy lies, and also, I don't agree this is a well worded question, because it has 2 solutions.

richwicks says

Oh - I think I see the problem - where do you come up with this equation:

a+b+c+d+b+c+d+c+d+d

?

to simplify, lets do

its not 2+4+4...its 24 + 4

richwicks says

I don't agree this is a well worded question, because it has 2 solutions.

I didn't see where it stated that there was only 1 solution

richwicks says

You must have misunderstood the question. It's asking:

ABCD +
 BCD +
  CD +
   D
----
6666

in my mind abcd = 6319 meaning a=6000, b=300, c=10 , d=9...bcd = 319, cd=19, d=9....total 6666

richwicks says

SunnyvaleCA - that Python?

Yes python. However, the asterisk symbol, designating multiplication in Phython, is treated as toggling bold text on and off due to the HTML "markdown" feature. I really am liking python for small tasks: very straightforward syntax, lots of examples out there on the internet, and fantastic built-in and 3rd party libraries.

As an example of 3rd party libraries... My girlfriend had a 500 page book with the binding cut off and a self-feeding scanner that produced PDF scans but only 1-sided. She wanted a single PDF with all the pages in order. I had her scan all the pages using the automatic sheet feeder, flip the stack upside down and scan them all again. By the time that was done I had a python script that used a free 3rd party PDF utilities library to reverse the second set of pages and then interleve them with the first set. Done!

stereotomy says

It's one equation with three unknowns. By the most elementary principles of linear algebra, this is a singular matrix, which necessarily has no unique solution.

Actually, it's 4 unknowns, but yeah. There are an infinite number of solutions except for the fact that you are constrained to integers between 1 and 9. That happens to limit it to 2 solutions.

Onvacation says

That's cheating! You're supposed to do it in your head!

I'm working smarter, not harder.

I believe everyone is overthinking it.

A=6000
B= 300
C= 20
D= 1.5

Answer:6321.5

Yep, I'll put $ on it.

Nobody said we needed to use base 10. How about bases 7 through 16. (Has to be at least base 7 because otherwise you can't write 6666 as the answer.

digits = "0123456789ABCDEF"
for base in range(6,16+1):
    for a,b,c,d in itertools.product(range(0,base),repeat=4):
        if a*base**3+b*base**2+c*base+d + b*base**2+c*base+d + c*base+d + d == (((6*base+6)*base+6)*base+6):
            print(base,
, digits[a],digits[b],digits[c],digits[d],
,a+b+c+d)

7  6 2 6 5  19
10  5 8 1 9  23
10  6 3 1 9  19
11  5 8 5 7  25
13  6 2 A 8  26
14  5 9 B 5  30
14  5 A 1 C  28
14  6 2 B 5  24
14  6 3 1 C  22

Interpreting the last line of output as an example: in base 14, the numbers for a,b,c,d are 6,3,1,C and the sum of those is 22 (in base 10).

^^^ the "pre" tag worked great. I also searched for * and replaced with &ast; .

komputodo says

richwicks says

Oh - I think I see the problem - where do you come up with this equation:

a+b+c+d+b+c+d+c+d+d

?

to simplify, lets do

its not 2+4+4...its 24 + 4

yes, and 2+4 = 6

It's adding the digit of each column together. What number do you think it ought to be?