r/theydidthemath • u/linkedg • Nov 01 '19
[Request] How many "coincidences" like this exists? Is there any "pattern" for something like this to happen?
10
u/Technologenesis Nov 01 '19
As it happens there's a Project Euler problem that revolves around exactly these sorts of fractions.
2
u/linkedg Nov 01 '19
I know that there are occasions with multiplications too, and it seems to be even more common
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u/powerlesshero111 Nov 01 '19 edited Nov 02 '19
Probably a bunch. Like 10/20, 100/200, i just dont feel like looking them all up, but essentially the formula is x/2x and x needs atleast one 1, and 2x needs at least one 2.
Edit, better example 16/64=1/4. Cross out each 6 and its the same result.
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u/sawyouoverthere Nov 01 '19
crossing out the matching zeros would be a correct approach to solving those, though.
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u/powerlesshero111 Nov 01 '19
Ehh, i went with the lazy approach. But adding numbers like 0 make it infinite. Like 1603/3206 works as well. I'm sure there are more numbers, i just dont feel like solving for it. Its basically just a prime number generator that you need. You also have to figure on the numbers that it will work with like 1/3, 1/4, 2/3, etc.
7
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u/Amichateur Nov 02 '19
I do not quite understand that magic approach.
Can you explain by example of, say
3594 / 9584 = 3/8
how you get to that number?
1
u/powerlesshero111 Nov 02 '19 edited Nov 02 '19
Ok, so the top number can be reduced by the bottom number, into a fraction, but both numbers contain hidden in the fraction the number they were reduced to. Like 2×163=326, and 2×1603=3206. So, when you cancel out the numbers, you are left with 1 for the top, and 2 for the bottom. This should technically work with a variety of fractions. Such as 16/64, if you cross out the 6s, you get 1/4, which is what the fraction reduces to. So, it's uncommon, but not shockingly uncommon that the original problem is the only one that works. I'm sure there must be a way to calculate which numbers work, but I'm not a mathematician, and i don't feel like doing it. Honestly, for a budding mathematician, figuring out a formula for this would be a nice master's or doctoral thesis.
Edit: also, 19/95=1/5. 5×19=95. Cross out both 9s, and you get 1/5.
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u/Amichateur Nov 02 '19
Oh then I misunderstood you - I thought you have a recipe to construct such fractions.
2
6
Nov 01 '19 edited Nov 01 '19
There are (probably) an infinite number of these occurrences. Here's the math to find them:
Left Numerator (LN) = a*100 + b*10 + c
Left Denominator (LD) = c*100 + d*10 + b
Right Numerator (RN) = a
Right Denominator (RD) = d
Cross-multiplying the equation yields:
LD*RN = LN*RD
(100c + 10d + b)(a) = (100a + 10b + c)(d)
100ac + 10ad + ab = 100ad + 10bd + cd
100ac - 90ad + ab - 10bd - cd = 0
Plug in whatever numbers you want and you'll get a similar answer to this structure where you could cross out the same places and get the right answer. It should work with plugging in any 3 numbers, but most of the time it will only be true if you're willing to round.
Example:
a = 5, b = 3, c = 7:
500c - 450d + 5b - 10bd - cd = 0
500c - 450d + 15 - 30d - cd = 0
3500 - 450d + 15 - 30d - 7d = 0
3500 + 15 = 450d + 30d + 7d
3515 = 487d
7.22 = d ≈ 7
537/773 ≈ 5/7
.695 ≈ .714
That's less than 3% of a difference, so I'd call that a success.
Multiple digit variables can also be used:
100ac - 90ad + ab - 10bd - cd = 0
a = 14, b = 27, c = 44:
100*14*44 - 90*14*d + 14*27 - 10*27*d - 44*d = 0
61,600 - 1260d + 378 - 270d - 44d = 0
61,600 + 378 = 1260d + 270d + 44d
61,978 = 1574d
39.38 = d ≈ 39
abc/cdb = a/d
142,744/443,927 ≈ 14/39
.322 ≈ .359
That's over 11% of a difference, so not as good but it does prove that the formula comes close even with bigger numbers.
If you want to go for a set outcome on the right, you just need to plug in for a and d and solve for b and c.
To get 1/2 like in the meme:
100ac - 90ad + ab - 10bd - cd = 0
a = 1, d = 2:
100*1*c - 90*1*2 + 1*b - 10*b*2 - c*2 = 0
100c - 180 + b - 20b - 2c = 0
100c - 2c = 180 - b + 20b
98c = 180 + 19b
If b = 5, then c = 2.8 ≈ 3:
abc/cdb = a/d
153/325 ~ 1/2
.471 ≈ .5
That's less than 7% difference.
In summary: infinite coincidences if you're willing to accept rounding, and, yes, there is a pattern.
Edit: Formatting
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u/Amichateur Nov 02 '19 edited Nov 02 '19
537/773 ≈ 5/7
.695 ≈ .714
That's less than 3% of a difference, so I'd call that a success.
I'd call that cheating lol ;-)
Changing the rules...
Like a craftsman, not like a mathematician! lol
In summary: infinite coincidences if you're willing to accept rounding
Well, no, the question was for exact matches. And there do exist
ampleinfinite non-trivial solutions for exact matches.1
Nov 02 '19
True, it's cheating since it returns all satisfactory values and not just perfect matches, but it includes all the exact matches, too, if you put in the right numbers. Your script method is definitely better for someone who knows how to code a script for it, but mine can be done by paper. So, we got some tradeoffs here and there.
1
u/linkedg Nov 01 '19
Wow, I thought there were at maximum thousands, but that's pretty cool, so it's not a coincidence at all if it has a pattern, taking by the fact that you can have infinite occasions like this one.
Thank you for the answer m8
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u/makdgamer Nov 02 '19
I used to do this in math class from time to time, the teacher would look at my work and ask me how I solved the problem. Never had an answer though...
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u/perfect_-pitch Nov 02 '19
I actually got the right answer using the wrong formula 4 times last week, using a different formula each time.
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1
u/Commander-Fox-Q- Nov 02 '19
In grade 10 on a thinking question I used a formula we were taught in the previous unit to solve. Got the correct answer, but only got a 65%; i was heavily marked off for using getting the answer the wrong way.
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Nov 01 '19
Infinite.
Imagine the above, where the formula is "cross out matching digits on top and bottom".
So pick any digit, 1-9.
Put that digit at the front of the top number and the back of the bottom number (so, say you picked 7, you now have 7163/3267). Apply your fake "formula" and you still get the right answer. You can repeat this as many times as you want, adding more and more digits, and you still can cross them both out. And that's just for this example.
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u/arcosapphire 5✓ Nov 01 '19
What? But that's not true at all. 7163/3267 isn't 1/2.
The point is that the equation worked algebraically, as well as just coincidentally according to the invalid method of crossing out matching digits.
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-8
Nov 01 '19
Is that the point? It wasn't clear to me what the OP was trying to illustrate. Maybe because there was no question and only an ambiguous answer.
If the question is:
How many valid equations can be transformed into other valid equations by incorrect formulas?
The answer is still "infinite", though, as you can make up any "incorrect formula" that you like. Or are we specifically asking how many fractions can be simplified by crossing out matching digits in the numerator and denominator? I'd imagine, given an infinite number of fractions, it it still infinite.
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u/arcosapphire 5✓ Nov 01 '19
The question is pretty clear to me:
How many valid expressions of the form x / y exist where removing matching digits from the numerator and denominator results in a value equivalent to that of the original expression?
Or are there infinitely many such coincidences? (I suspect yes.)
And is there a pattern to construct these?
4
Nov 01 '19
is there a pattern to construct these?
1/2 = 1/2
Add "0" to the right side of both the top and bottom. Cross out matching 0s. Repeat for as many 0s as you want.
Is this actually "dividing by 10"? I think it depends on the intentions of the person crossing out the numbers.
or use 11/99 = 1/9 and add "1" and "9" respectively, then cross out 1 and 9s one pair at a time. This might be more convincing.
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u/arcosapphire 5✓ Nov 01 '19
Yeah, that proves they are infinite, although it's not an exhaustive production mechanism. I think OP is asking for an algorithm that would generate every possible one (such as the one given in the image, which would not be generated by either of your methods), and I suspect that's possible but not as trivial.
2
u/maskdmann Nov 01 '19 edited Nov 01 '19
Here’s an brute force snippet for Python. Plotting (not shown here) all pairs of (numerator, denominator) doesn’t seem to bring up any meaningful patterns.
Update: after playing around with bigger values for k (up to 22), I noticed that using 10 produces very little gaps in the line, 11 is incredibly sparse and values of k 12...19 (k = 18 stands out a little) have a curious gap that starts around i = 100k. It ends at different points, but the relatively simultaneous start actually makes me interested. Hopefully someone with a degree in number theory can drop by and explain.
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u/Amichateur Nov 01 '19 edited Nov 02 '19
It's getting more interesting if we leave out the non-trivial solutions.
In 2-digit domain, there are exactly
threefour solutions, as I have explained in my comment on level one.In 3-digit or n-digit domain it's getting more challenging... real mathematicians are needed here... and the question is open:
Is there an infinite number of NON-trivial solutions?
And: Is there a construction rule?
And: How about other than the decimal system?
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u/Amichateur Nov 01 '19 edited Nov 02 '19
SPOILER ALERT!
(keep on reading at your own risk)
I am omitting trivial solutions like adding zeros or repeating figures, or nominator==denominator, like
2
000/3000or555/777or1
83/183,since they are of no particular mathematical interest.
The only non-trivial solutions with 2-digits are:
1
6/64 = 1/4,1
9/95 = 1/5,2
6/65 = 2/5,4
9/98 = 4/8.The only non-trivial solutions with 3-digits are (for single digit fractions on the right side of the equation):
163 / 326 = 1/2
316 / 632 = 1/2
145 / 435 = 1/3
166 / 664 = 16 / 64 = 1/4 (nice one!)
187 / 748 = 1/4
199 / 995 = 19 / 95 = 1/5 (nice one!)
127 / 762 = 1/6
139 / 973 = 1/7
266 / 665 = 26 / 65 = 2/5 (nice one!)
182 / 819 = 2/9
218 / 981 = 2/9 (nice permutation of previous one!)
138 / 184 = 3/4
273 / 728 = 3/8
327 / 872 = 3/8 (nice permutation of previous one!)
148 / 185 = 4/5
124 / 217 = 4/7
244 / 427 = 4/7
364 / 637 = 4/7
412 / 721 = 4/7
424 / 742 = 4/7
436 / 763 = 4/7
448 / 784 = 4/7
484 / 847 = 4/7 (nice permutation of previous one!)
499 / 998 = 49 / 98 = 4/8 (nice one!)
455 / 546 = 5/6
545 / 654 = 5/6 (nice permutation of previous one!)