Phil Carmody <pc+usenet@asdf.org> wrote:

"Keith F. Lynch" <kfl@KeithLynch.net> writes:

HenHanna <HenHanna@dev.null> wrote:

Keith F. Lynch wrote:

Each of them has a sum that's equal to its product and is an integer.

They're literally everywhere. Given 2 rational numbers, there's a solution

(Here I've capped numerators and denominators to 50.)

You missed the "and is an integer" part.

I'm particularly enamoured with this find:
4/3 7/6 9/2 7=7

As well you might, since that was the only solution that met my
criteria. But of course it was one of the solutions I listed
in the first place.

Again, I wish you luck in finding one that I didn't list.
Even though there are infinitely many such.
--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.

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On Wed, 11 Sep 2024 12:15:12 -0000 (UTC), Keith F. Lynch wrote:

49/15, 25/21, 54/35

--
David Entwistle

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On Wed, 11 Sep 2024 12:15:12 -0000 (UTC), Keith F. Lynch wrote:

I discovered that these three sets of three positive rationals have an interesting property in common:

52/39, 7/6, 9/2

--
David Entwistle

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"Keith F. Lynch" <kfl@KeithLynch.net> writes:

HenHanna <HenHanna@dev.null> wrote:

Keith F. Lynch wrote:

Since it's been more than a week, and nobody has figured it out:
Each of them has a sum that's equal to its product and is an integer.

i think one person said exactly that.

Who and when? I didn't see any such post.

Is it easy to find them?

No, even though there are infinitely many. Try and find one I
didn't list.

Constraints: All three numbers must be positive, real, and rational,
but not integers.

They're literally everywhere. Given 2 rational numbers, there's a solution

(Here I've capped numerators and denominators to 50.)

$ echo 'for(n=1,10000,a=(1+random(49))/(2+random(48));if(denominator(a)==1,next);b=1/a+(1+random(50))/(2+random(50));if(denominator(b)==1||numerator(b)>=50,next);x=(a+b)/(a*b-1);if(denominator(x)>1&&numerator(x)<50&&denominator(x)<50,print(a" "b" "x" "a+
b+x"="a*b*x)))' | gp -q > crap
$ sort -n crap | uniq
1/2 11/4 26/3 143/12=143/12
1/2 16/5 37/6 148/15=148/15
1/2 19/8 46/3 437/24=437/24
1/2 26/3 11/4 143/12=143/12
1/2 29/7 13/3 377/42=377/42
1/2 8/3 19/2 38/3=38/3
1/3 21/5 34/3 238/15=238/15
1/3 25/7 41/2 1025/42=1025/42
1/3 31/5 49/8 1519/120=1519/120
1/3 36/7 23/3 92/7=92/7
1/4 11/2 46/3 253/12=253/12
2/3 11/2 37/16 407/48=407/48
2/3 11/6 45/4 55/4=55/4
2/3 7/2 25/8 175/24=175/24
2/3 9/4 35/6 35/4=35/4
2/5 7/2 39/4 273/20=273/20
3/2 13/7 47/25 1833/350=1833/350
3/2 17/3 43/45 731/90=731/90
3/2 29/24 10/3 145/24=145/24
3/2 3/2 12/5 27/5=27/5
3/2 4/3 17/6 17/3=17/3
3/2 5/3 19/9 95/18=95/18
3/2 8/3 25/18 50/9=50/9
3/4 19/3 17/9 323/36=323/36
3/4 8/3 41/12 41/6=41/6
3/5 11/3 32/9 352/45=352/45
4/3 12/11 16/3 256/33=256/33
4/3 13/2 47/46 611/69=611/69
4/3 13/4 11/8 143/24=143/24
4/3 17/6 3/2 17/3=17/3
4/3 49/12 39/32 637/96=637/96
4/3 7/4 37/16 259/48=259/48
4/3 7/6 9/2 7=7
4/3 9/4 43/24 43/8=43/8
4/5 7/3 47/13 1316/195=1316/195
5/2 7/5 39/25 273/50=273/50
5/3 8/5 49/25 392/75=392/75
6/5 3/2 27/8 243/40=243/40
6/7 19/12 41/6 779/84=779/84
6/7 9/2 15/8 405/56=405/56
7/2 5/3 31/29 1085/174=1085/174
7/3 9/7 38/21 38/7=38/7
7/9 29/7 31/14 899/126=899/126
8/5 3/2 31/14 186/35=186/35
9/2 5/3 37/39 185/26=185/26
15/7 3/5 48/5 432/35=432/35
20/21 5/2 5/2 125/21=125/21
22/7 10/11 24/11 480/77=480/77
42/11 20/21 38/21 1520/231=1520/231

I'm particularly enamoured with this find:
4/3 7/6 9/2 7=7

Widening the net, I was unable to find any other integer sum/products.
Can anyone else find another one - I suspect actually using some algebra
might make sense, rather than just probing randomly.

Other small denominators are everywhere:

1/12 63/4 152/3 133/2=133/2
1/28 147/4 824/7 309/2=309/2
1/3 9/2 29/3 29/2=29/2
1/33 729/22 24079/3 16119/2=16119/2
1/5 29/5 75/2 87/2=87/2
1/7 49/2 69/7 69/2=69/2

1/15 78/5 1175/3 1222/3=1222/3
1/18 172/9 621/2 989/3=989/3
1/2 16/3 7/2 28/3=28/3
1/2 19/2 8/3 38/3=38/3
1/2 7/2 16/3 28/3=28/3
1/4 16/3 67/4 67/3=67/3
1/4 19/4 80/3 95/3=95/3
1/5 25/3 64/5 64/3=64/3
1/5 64/5 25/3 64/3=64/3
1/6 15/2 92/3 115/3=115/3
1/6 20/3 123/2 205/3=205/3
1/7 55/7 196/3 220/3=220/3
1/78 169/2 13184/13 3296/3=3296/3
1/8 103/8 64/3 103/3=103/3
2/15 25/3 381/5 254/3=254/3
2/7 26/7 196/3 208/3=208/3
3/2 17/6 4/3 17/3=17/3
3/2 28/3 5/6 35/3=35/3
3/2 5/6 28/3 35/3=35/3
4/3 3/2 17/6 17/3=17/3
4/3 39/2 5/6 65/3=65/3
4/3 5/6 39/2 65/3=65/3
5/6 3/2 28/3 35/3=35/3
5/6 4/3 39/2 65/3=65/3




Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have gained some understanding of the world in which we live. As such, we can cast aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/

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I discovered that these three sets of three positive rationals have an interesting property in common:

9/2, 4/3, 7/6

49/15, 25/21, 54/35

49/2, 4/7, 27/14

If nobody figures it out, I will provide the answer in a week.
--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.

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Keith F. Lynch <kfl@KeithLynch.net> wrote:

I discovered that these three sets of three positive rationals have
an interesting property in common:

9/2, 4/3, 7/6
49/15, 25/21, 54/35
49/2, 4/7, 27/14

If nobody figures it out, I will provide the answer in a week.

Since it's been more than a week, and nobody has figured it out:

Each of them has a sum that's equal to its product and is an integer.

For instance 9/2 + 4/3 + 7/6 = 7 and 9/2 x 4/3 x 7/6 = 7.

Here are some more triples with this same unusual property:

121/42, 637/66, 36/77
81/5, 50/9, 11/45
625/18, 81/50, 148/225
289/15, 950/51, 9/85
49/3, 207/7, 2/21
450/13, 169/15, 23/195
25/2, 252/5, 1/10
81/2, 292/9, 1/18
242/5, 325/11, 3/55
121/2, 92/11, 3/22
625/21, 4214/75, 81/1575
676/7, 49/26, 99/182
245/3, 198/7, 1/21
343/3, 81/7, 2/21
578/5, 175/17, 9/85
529/3, 126/23, 13/69
289/2, 756/17, 1/34
525/2, 52/5, 1/10
--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.

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On Wed, 18 Sep 2024 22:39:45 +0000, Keith F. Lynch wrote:

Keith F. Lynch <kfl@KeithLynch.net> wrote:

I discovered that these three sets of three positive rationals have
an interesting property in common:

9/2, 4/3, 7/6
49/15, 25/21, 54/35
49/2, 4/7, 27/14

If nobody figures it out, I will provide the answer in a week.

Since it's been more than a week, and nobody has figured it out:

Each of them has a sum that's equal to its product and is an integer.

i think one person said exactly that.


Is it easy to find them?



How about 2 numbers

2+2 = 2*2

1+x = 1*x ------- No sol.

3+x = 3*x ------- easy to solve.

A+x = A*x ------- all easy to solve.

For instance 9/2 + 4/3 + 7/6 = 7 and 9/2 x 4/3 x 7/6 = 7.

Here are some more triples with this same unusual property:

121/42, 637/66, 36/77
81/5, 50/9, 11/45
625/18, 81/50, 148/225
289/15, 950/51, 9/85
49/3, 207/7, 2/21
450/13, 169/15, 23/195
25/2, 252/5, 1/10
81/2, 292/9, 1/18
242/5, 325/11, 3/55
121/2, 92/11, 3/22
625/21, 4214/75, 81/1575
676/7, 49/26, 99/182
245/3, 198/7, 1/21
343/3, 81/7, 2/21
578/5, 175/17, 9/85
529/3, 126/23, 13/69
289/2, 756/17, 1/34
525/2, 52/5, 1/10

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XPost: sci.lang

On 9/11/2024 5:15 AM, Keith F. Lynch wrote:

I discovered that these three sets of three positive rationals have an interesting property in common:

9/2, 4/3, 7/6

49/15, 25/21, 54/35

49/2, 4/7, 27/14

If nobody figures it out, I will provide the answer in a week.

that sounds good.

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Keith F. Lynch used his keyboard to write :

I discovered that these three sets of three positive rationals have an interesting property in common:

9/2, 4/3, 7/6

49/15, 25/21, 54/35

49/2, 4/7, 27/14

If nobody figures it out, I will provide the answer in a week.

normalizing to lowest common denominator I get

27/6 8/6 7/6
343/105 125/105 162/105
343/14 8/14 27/14

where the numerator of first and second is a cube and the third is not
:-)

--
/-\ /\/\ /\/\ /-\ /\/\ /\/\ /-\ T /-\
-=- -=- -=- -=- -=- -=- -=- -=- - -=-
........... [ al lavoro ] ...........

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HenHanna <HenHanna@dev.null> wrote:

Keith F. Lynch wrote:

Since it's been more than a week, and nobody has figured it out:
Each of them has a sum that's equal to its product and is an integer.

i think one person said exactly that.

Who and when? I didn't see any such post.

Is it easy to find them?

No, even though there are infinitely many. Try and find one I
didn't list.

Constraints: All three numbers must be positive, real, and rational,
but not integers.

And of course have to be in simplest form, i.e. 1/2, not 2/4. One
person posted "52/39, 7/6, 9/2" which is of course the same three
numbers as my "9/2, 4/3, 7/6". And he didn't say what property
they had, anyway.

Without any constraints, x,i,-i is always a solution for any x,
if i is the square root of minus one.

How about 2 numbers

Too simple to be interesting. If you want two numbers to have a sum
of S and a product of P, whether or not S=P, the two numbers will be
S + sqrt(S^2 - 4P) and S - sqrt(S^2 - 4P).
--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.

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XPost: sci.lang

Keith F. Lynch wrote:

Since it's been more than a week, and nobody has figured it out:
Each of them has a sum that's equal to its product and is an integer.

i think one person said exactly that.

----------- Who and when? I didn't see any such post.


So this post below didn't get to your site (or Newsreader).




On Fri, 13 Sep 2024 12:58:08 +0000, IlanMayer wrote:

On Fri, 13 Sep 2024 2:19:31 +0000, HenHanna wrote:

On 9/11/2024 5:15 AM, Keith F. Lynch wrote:

I discovered that these three sets of three positive rationals have an
interesting property in common:

9/2, 4/3, 7/6

49/15, 25/21, 54/35

49/2, 4/7, 27/14

If nobody figures it out, I will provide the answer in a week.

that sounds good.

SPOILER

The sum of each triplet is the same as its product.

9/2*4/3*7/6 = 9/2+4/3+7/6 = 7
49/15*25/21*54/35 = 49/15+25/21+54/35 = 6
49/2*4/7*27/14 = 49/2+4/7+27/14 = 27

Please reply to ilanlmayer at gmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

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XPost: sci.lang

On Fri, 13 Sep 2024 2:19:31 +0000, HenHanna wrote:

On 9/11/2024 5:15 AM, Keith F. Lynch wrote:

I discovered that these three sets of three positive rationals have an
interesting property in common:

9/2, 4/3, 7/6

49/15, 25/21, 54/35

49/2, 4/7, 27/14

If nobody figures it out, I will provide the answer in a week.

that sounds good.

SPOILER
































The sum of each triplet is the same as its product.

9/2*4/3*7/6 = 9/2+4/3+7/6 = 7
49/15*25/21*54/35 = 49/15+25/21+54/35 = 6
49/2*4/7*27/14 = 49/2+4/7+27/14 = 27

Please reply to ilanlmayer at gmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

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XPost: sci.lang

HenHanna <HenHanna@dev.null> wrote:

So this post below didn't get to your site (or Newsreader).
IlanMayer wrote:

. . .
The sum of each triplet is the same as its product.

9/2*4/3*7/6 = 9/2+4/3+7/6 = 7
49/15*25/21*54/35 = 49/15+25/21+54/35 = 6
49/2*4/7*27/14 = 49/2+4/7+27/14 = 27

Please reply to ilanlmayer at gmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

Thanks for forwarding that. I don't know why that post never showed
up on Panix.
--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.

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XPost: sci.lang

Op 13-9-2024 om 04:19 schreef HenHanna:

On 9/11/2024 5:15 AM, Keith F. Lynch wrote:

I discovered that these three sets of three positive rationals have an
interesting property in common:

9/2, 4/3, 7/6

49/15, 25/21, 54/35

49/2, 4/7, 27/14

If nobody figures it out, I will provide the answer in a week.

        that sounds good.

They are three, they are positive, they are rationals, they're written
in the same form and the same number base or whatsitcalled, they are
mentioned here (and no others are).
Anything else yet?

--
guido wugi :o)

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