interesting task
John Smith
a1, a2, ..., an, b1, b2, ..., bn
satisfying conditions:
a1 + a2 + ... + an = b1 + b2 + ... + bn
and
a1 * a2 * ... * an = b1 * b2 * ... * bn
any advice/tips?
thanks
Gerry Myerson
<835538205.51444.12594...@gallium.mathforum.org>,
John Smith <j0hns...@yahoo.com> wrote:
Some funny symbols there. I take it what's wanted is two sets of
positive integers with the same sum and the same product. I suspect
this can be done for all n > 2, as follows:
Given n > 2, choose some number N with lots and lots and lots
of factors, write down all the many many ways of writing N as
a product of n naturals, and most likely you'll find two that have
the same sum.
For n = 3 an example (maybe the smallest example?)
is 3 + 8 + 10 = 4 + 5 + 12 = 21,
3 x 8 x 10 = 4 x 5 x 12 = 240.
But looking at ways to factor 360 yields 5 solutions for n = 3
(11, 12, 30 and 2, 5, 36
2, 12, 15 and 3, 6, 20
3, 10, 12 and 4, 6, 15
4, 9, 10 and 5, 6, 12
5, 8, 9 and 6, 6, 10)
and it's that plethora of solutions that makes me confident
that if you choose N with enough factors you'll be able to find
two sets of n with the required property.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
John Smith
Gerry Myerson
<414950066.56270.12595...@gallium.mathforum.org>,
John Smith <j0hns...@yahoo.com> wrote:
> is there a possibility to prove in a clearly way that the task can be done
> for all n>2?
What task? You've snipped all the context.
bill
What are "pairwise distinct positive integers"? Does this mean that
all the numbers are different? No duplicates? a= 1,2,9 and
b = 2,3,7 would not be allowed?
regards, Bill J
Gerry Myerson
<3a90bea2-0e42-490b...@a21g2000yqc.googlegroups.com>,
bill <b92...@yahoo.com> wrote:
Surely it means you must avoid a_i = b_j,
otherwise as soon as you have a solution for some value of n
you trivially get solutions for all bigger n.
Possibly it also means you must have a_i = a_j implies i = j,
and b_i = b_j implies i = j.
Leon Aigret
wrote:
>Find all naturals n for which there exist pairwise distinct positive integers
>a1, a2, ..., an, b1, b2, ..., bn
>satisfying conditions:
>a1 + a2 + ... + an = b1 + b2 + ... + bn
>and a1 * a2 * ... * an = b1 * b2 * ... * bn
>
>any advice/tips?
>thanks
and on Mon, 30 Nov 2009 06:11:17 EST, he continued with:
>is there a possibility to prove in a clearly way that the task can be done for all n>2?
There are simple ways to derive new solutions from old ones.
If a1, ..., an plus b1, ..., bn is a solution, then so is any
scaled version k * a1, ..., k * an plus k * b1, ..., k * bn. Also, if
a1, ..., an1 plus b1, ..., bn1 and c1, ..., cn2 plus d1, ..., dn2
are solutions for n1 and n2 without common entries, then
a1, ..., an1, c1, ..., cn2 plus b1, ..., bn1, d1, ..., dn2
is a solution for n1 + n2. Solutions with common entries can always be
avoided by scaling one original solution out of the danger zone, so
solutions for n1 and n2 always give a solution for n1 + n2 and Gerry
Myerson's solution for n = 3 also produces solutions for n = 6, 9, ...
Combinations that also use the solution 1, 10, 15, 4 plus 5, 2, 3, 20
for n = 4 can be used to fill in the gaps. This only leaves the case
n = 5 unsolved.
Leon
Gerry Myerson
aigre...@myrealbox.com (Leon Aigret) wrote:
3, 18, 22, 24, 26 and 9, 11, 12, 13, 48.
Another way to do it without scaling is to note that for all a and b,
a + (b + 1) + (2 a + 2) + 2 b = (a + 1) + b + 2 a + (2 b + 2) and
a (b + 1) (2 a + 2) (2 b) = (a + 1) b (2 a) (2 b + 2)
so once you have solutions for n = 3, 4, 5, and 6 you can tack on
solutions for appropriate values of a and b to increase n by 4
and eventually get every n. Of course, n =4 is immediate with, say,
a = 2, b = 7 giving 2, 6, 8, 14 and 3, 4, 7, 16.
Gerry Myerson
Here's another way to do things. For n at least 3,
there's always a solution where one set is of the form
a_1, a_2, ..., a_{n-1}, 2^{n-1} a_n
and the other is of the form
2 a_1, 2 a_2, ..., 2 a_{n-1}, a_n.
All that's needed is to have
a_1 + a_2 + ... + a_{n-1} = (2^{n-1} - 1) a_n.
For n > 3, you can take a_n = n - 1, and take a_1, a_2, ..., a_{n-1}
to be numbers on either side of 2^{n-1} - 1.
E.g., for n = 4, we have 2^{n-1} - 1 = 7, so we take a_1, a_2, a_3
to be 6, 7, 8, and a_4 = 3, yielding
6, 7, 8, 24 vs 12, 14, 16, 3.
For n = 5, we have 2^{n-1} - 1 = 15,
so we take a_1, a_2, a_3, a_4 to be 13, 14, 16, 17, and we get
13, 14, 16, 17, 64 vs 26, 28, 32, 34, 4.
So for each n > 3 this gives a solution which does not depend on knowing
solutions for smaller n.
I hope this wasn't someone's homework.
Bill Taylor
but you can also have mixed-sign solutions...
-15 2 7 and -14 3 5 .
................
Amusing solutions (for n = 3) occur when a number from
one set is the product of two from the other, and vice versa.
e.g. 25 9 3 v 27 5 5 and 20 7 3 v 21 5 4
and 12 7 2 v 14 4 3 etc.
I don't know if these solutions would be considered
better or worse than the "more random" ones.
................
The whole puzzle has a strong smell of the similar puzzle
whereby one must find SINGLE sets of numbers whose sum
equals their product. It has appeared here occasionally.
[ e.g. 2 2 or 3 2 1 or 4 2 1 1 etc]
The two problems are completely different, OC,
but clearly from the same stable.
-- Wittering William
** Dogs believe that they are Men.
** Cats believe that they are God.
Gerry Myerson
<5ca9f8ab-afc8-4dc1...@r24g2000yqd.googlegroups.com>,
Bill Taylor <w.ta...@math.canterbury.ac.nz> wrote:
> I know it's outside the terms of the original enquiry,
> but you can also have mixed-sign solutions...
>
> -15 2 7 and -14 3 5 .
>
> ................
>
> Amusing solutions (for n = 3) occur when a number from
> one set is the product of two from the other, and vice versa.
>
> e.g. 25 9 3 v 27 5 5 and 20 7 3 v 21 5 4
> and 12 7 2 v 14 4 3 etc.
> I don't know if these solutions would be considered
> better or worse than the "more random" ones.
One can find a general formula for these.
You want a + b + c d = a b + c + d,
which amounts to (a - 1) (b - 1) = (c - 1) (d - 1),
so you can take any number which has two different factorizations,
e.g., 100 = 5 x 20 = 4 x 25
and turn it into 6, 21, 130 vs 5, 26, 126.
So it's r s + 1, t u + 1, (r t + 1)(s u + 1)
vs r t + 1, s u + 1, (r s + 1)(t u + 1).