[Request] Got this text today. Where do I even start?

[Request] Got this text today. Where do I even start?

You can also cheat after you figure out the first digit is 5:

The combination repeats itself three times. If you know that the total is 555, and there are 3 identical numbers, 555/3 = 185

The circle with horizontal stripes is 1, vertical striped circle is 8, the dot is 5.

How I figured it:

Started at the ones position: Some number times 3 must yield itself in the ones position (( 3 * x) % 10) = x. The only two numbers that satisfy this are 0 or 5. It can't be 0 because the equation can't work.

The tens position: some number times 3 plus the carried 1 from the ones column (3 * y + 1) % 10) = 5. The only digit that satisfies this is 8.

The hundreds position: some number times 3 plus the carried 2 from the tens column: (3 * z + 2) % 10 = 5. The only number that satisfies this is 1.

That's working smart, not hard.

Kind of ashamed to admit I did it the brute force way now.

Brute force is satisfying in itself.

Wow that is a lot simpler than what I was trying to do. Thanks! ✓

Working smarterer, not harder:

>>> from itertools import permutations >>> for a in permutations('0123456789', 3): ... if int(''.join(a)) * 3 == int(a[2] + a[2] + a[2]): ... print a ... ('1', '8', '5')

First, I'm going to assume everything is either zero or a positive, single-digit integer, for simplicity.

Change everything to Xs and Ys first of all. Shapes are hard to write down. Remember too that XYZ are not multiplicative, but are simply digits. There are no operational properties explaining them.

XYZ

XYZ

+XYZ

=ZZZ

So we know that 3(XYZ)=ZZZ.

Now, let's try a few cases. What we want to do is look at what happens when we add the Z's in the right column. In base 10, Where does 3Z=Z? We know that the answer is either 0 or 5.

Case 1: Z=0

XY0 + XY0 + XY0 = 000

(I'm skipping a step here in the proof)

therefore XY + XY+ XY=00. we know that with positive integers, the only possible solution is X=0, Y=0.

So Answer: X=0, Y=0, Z=0

Case 2: Z=5

So we have XY5+XY5+XY5=555

Or 3(XY5)=555

XY5=555/3

XY5=185

X=1, Y=8, Z=5

Well, there's nothing that says the symbols are unique, so it could just be 0,0,0

Take all numbers created by repeating a number three times.

000

111

222

333

etc...

Include

101010

111111

121212

because they said number, not digit.

Divide each number by three.

000/3=0

111/3=37

222/3=74

etc...

The ones that contain the repeating part at the end are acceptable answers.

000+000+000=000

185+185+185=555

168350+168350+168350=505050

166833500+166833500+166833500=500500500

166683335000+166683335000+166683335000=500050005000

etc...

The 185 answer was probably what they had in mind, but they said numbers, not digits.

There's also nothing that says the symbols are from a base-10 system.

The base (b) has to be divisible by two so that the last digit remains the same. The last digit has to be half of the base, and can't be divisible by 3, so that all the numbers are different (ex. in base 12 666/3 = 222). Also, the repeated-digit number has to be divisible by 3. [(b3 * b/2) + (b2 * b/2) + b/2] / 3 has to be an integer.

I checked all the bases up to 30. Only 4 work:

In base 10, the solution is 185.

In base 16, it's 2d8.

In base 22 it's 3ib.

In base 28, it's 4ne.

It seems like it works every 6th base after 10. I'm not sure why.

So with the help of /u/Whitt83's solution for base 10, my friend and I were able to find a general solution for all bases. Sadly, I'm not particularly skilled at proofs, so this is not 100% proven, but it definitely seems right.

For simplicity, let's call the circle with horizontal stripes A, the one with vertical stripes B, and the little circle C.

The general solution is for a natural number N, the base M = 6N + 4 ; A = N ; B = 5N + 3 ; C = 3N + 2.

The first thing I did is use the generalization that C*3 must end in C, and figure out for the first few bases what numbers work for that. To start, it's clear that if a digit is less than or equal to M/3, it cannot work, because to "loop around" to its own digit, it must both surpass that base's "10," and reach its own digit. If the digit is M/3, it will equal 10, and if it's less than that, it will be larger, but not surpass 10. Additionally, 0 is an answer for all solutions, but that's boring so I'll ignore it.

For base 2, 1 works (1+1+1=11). Base 3, nothing (13=10 ; 23=20). Base 4, 2 and 3 (13=3 ; 23=12 ; 33=21) Base 5, nothing (23=11 ; 33=14 ; 43=22). Base 6, 3 (33=13 ; 43=20 ; 53=23). Base 7, nothing (33=12 ; 43=15 ; 53=21 ; 6*3=24). The clear pattern is that only even bases have a digit that will repeat itself in the 1's place if multiplied by 3, and that that digit is half the value of the base. Notation wise, M must be even, and C = M/2.

To show how I'm going to do the rest of the "proof," here's an example of how to solve for a specific base. I'm doing base 4, for reasons that will be clear later: C=M/2=2. Therefore, the final number is CCC=2224 (base 4). We need to divide this by 3 to get the original number we added together 3 times (i.e. multiplied by 3). 2224/3=324. We know this works because the digits in the 1's slot for both number is the same. In this case, A=0, B=3, C=2.

Now let's try to narrow it down more. When the M is divisible by 3, that means that, since it's also divisible by 2, C is divisible by 3 (M=6, so C=3; M=12, so C=6; M=18, so C=9). For the example of M=6, C=3, so the final number is 3336. 3336/3 = 1116. Wait. Shit. 1≠3. This must mean that 6 doesn't work as a base. To generalize, if m is divisible by 3 and even, C=M/2, and the number is CCCM. When we divide it by 3, we get CCCM/3. The number ends in m/2, and since m is divisible by both 3 and 2, it must be a positive integer, and notably not C. Therefore, m cannot be divisible by 3.

Lastly, let's just go with trial and error for a few more options.

Base 2: 1112 is not divisible by 3 (710/3) Base 4: 2224/3 = 324 Base 8: 4448 is not divisible by 3 (=29210/3) Base 10: 555/3 = 185 Base 14: 77714/3 is not divisible by 3 (=147710/3) Base 16: 88816/3 = 2d816 (d=13) Base 20: aaa20 (a=10) is not divisible by 3 (=421010/3) Base 22: bbb22/3 = 3ib22 (b=11, i=18)

SO, that clearly shows that when m = 6n + 2, CCC is not divisible by 3, meaning that those do not have any solution.

Finally, to find generalizations for each value, let's look at M, A, B, and C's values for each m that works.

M A B C 4 0 3 2 10 1 8 5 16 2 13 8 22 3 18 11 28 4 23 14

And so on. Since I've already said what the rule is for each value at the beginning, and in the spirit of this post, here are the solutions again, in base 7, where a=0, b=1, ..., f=5, g=6

In general, for a given natural number N, base M = gN + e ; A = N ; B = fN + d ; and C = dN + c such that ABC + ABC + ABC = CCC where A, B, and C are digits.

EDIT: Like I said above, this is only a "proof" by example, not a thorough proof, so feel free to jump off of this and more thoroughly prove what are currently conjectures.

Confirmed: 1 request point awarded to /u/Whitt83. [History]

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Technically correct: the best kind of correct.

Assuming the symbols are graphical representations of decimal numbers, we start by defining the problem in more algebraic terms:

+(100A + 10B + C)

+(100A + 10B + C)

+(100A + 10B + C)

= (100C + 10C + C)

Or more simply: 300A+30B+3C = 111C

Solving for B, this becomes: B = 18C/5-10A

Since we know these values have to be positive integers between 0 and 9, that fraction means C is either 0 or 5. However, if C = 0, then either A or B would have to be negative, so C = 5. This leaves us with: B = 18-10A.

The only value for A that yields a positive, single-digit, integer value for B is 1, therefore A = 1, and B = 8.

And indeed, 185*3 = 555.

They also didn't specify the base, or even say that the symbols represent different numbers.

Thanks! You did it a different way than u/Whitt83 and got the same answer. This is how I started out but couldn't figure out what to do after figuring out z=5. ✓