Large Integer

Originally posted by Palynka
Thanks!

Do you have a definite proof?

Suppose 1/a+1/b+1/c+1/d=1 ; then the largest number of the same format which is less than 1 is: 1/a+1/b+1/c+1/(d+1) and then (1/d)-(1/(d+1))=1/E...
. Like wise, if 1/a+1/b+1/c=1, then (1/c)-(1/(c+1))=d...So if we knew c, then we would know d and therefore E as well. But c can only be 6. and so d=42 and E=1806.

Originally posted by ThudnBlunder
Suppose 1/a+1/b+1/c+1/d=1 ; then the largest number of the same format which is less than 1 is: 1/a+1/b+1/c+1/(d+1) and then (1/d)-(1/(d+1))=1/E...
. Like wise, if 1/a+1/b+1/c=1, then (1/c)-(1/(c+1))=d...So if we knew c, then we would know d and therefore E as well. But c can only be 6. and so d=42 and E=1806.

That was my reasoning, but maybe I'm nitpicking but I don't think that's a proof.

Sure, let's say I have 1/a+1/b+1/c+1/d=1 and do like you said and find a solution. But I can also find 1/e+1/f+1/g+1/h=1 and also find another solution. For example:

1/2+1/3+1/8+1/24 = 1/2+1/3+1/7+1/42 = 1.

What makes me sure that proceeding iteratively (from 1 element to 5) will get me the unique solution? We need a proof for that...

Maybe we can prove that there is no combination of 1/A + 1/B + 1/C + 1/D which gets closer to 1 (but still below it) than the series produced by our function

Originally posted by Palynka
That was my reasoning, but maybe I'm nitpicking but I don't think that's a proof.

Sure, let's say I have 1/a+1/b+1/c+1/d=1 and do like you said and find a solution. But I can also find 1/e+1/f+1/g+1/h=1 and also find another solution. For example:

1/2+1/3+1/8+1/24 = 1/2+1/3+1/7+1/42 = 1.

What makes me sure that proceeding iteratively (from 1 element to 5) will get me the unique solution? We need a proof for that...

I think I understand what you mean. My "proof" was a bit lazy. I could try to make it more rigorous, but maybe an easier way to see the proof is to think of it this way:

We want the smallest possible value(largest denominator) for 1/E. Since we require the fifth term to be as small as possible, the sum of the first four terms must be as large as possible. Not quite so obvious perhaps, but if the fifth term is to be as small as possible, the fourth, 1/D must be as large as possible, without making the fifth, 1/E unnecessary; in other words without making the total 1 after only four terms or elements. Also the third 1/C must be as large as possible without making the fourth unnecessary. At each stage the total has to be as close as possible to 1, but only reaching 1 at the nth term. The first two terms cannot make the total 1, so they are always 1/2 and 1/3. From the third term onwards, adding 1 to the denominator makes the total as large as possible while allowing for one more term to be added.

Your example 1/2+1/3+1/8+1/24 =1, fails because you have not made the third term as large as possible, 1/7. I hope it's clearer now.

Originally posted by ThudnBlunder
What is the largest integer E, such that 1/A+1/B+1/C+1/D+1/E=1 ??
A<B<C<D<E are all positive integers

1/e = 1 - 1/a - 1/b - 1/c - 1/d
= 1 - (bcd + acd + abd + abc)/abcd

1/e = (abcd - (bcd + acd + abd + abc))/abcd

as the top is an integer, 1/e cannot be smaller than 1/abcd

consider our solution:
A = 2
B= A+1
C= BA+1
D= CBA+1
E= DCBA

1/E = 1/(A*B*C*D)

so we have proved our function makes 1/E is as small as it can be!

oops, no, because I haven't proved that abcd is as large as it can be 🙁

Originally posted by iamatiger
oops, no, because I haven't proved that abcd is as large as it can be 🙁

Yes, that seems to be the most difficult part of the proof. Rigorous, formal proof is not my strong suit, but here is my best attempt.

We want the smallest possible fraction for 1/E; such that:

1)).........1/A+1/B+1/C+1/D+1/E=1 ;
with A<B<C<D<E ..all integers.

To make 1/E as small as possible, we need the sum of the first four to be as large as possible. Suppose: that 1/P is the smallest possible fraction such that :

2))........1/M+1/N+1/O+1/P=1

If 1/P is as small as it can be, then the sum of the first three is as large as it can be. Clearly, then, the largest sum we can get (less than 1) with only four terms is :

1/M+1/N+1/O+1/(P+1) So:

3))........1/M+1/N+1/O+1/(P+1)+1/E=1

Subtract 3)) from 2))

4))........1/P-1/(P+1)=1/E

Because we have assumed that 1/P is as small as possible, the sum of the first three fractions must be as large as possible.
Now suppose 1/Z is the smallest value such that :

5)).......1/X+1/Y+1/Z=1

The largest sum(less than 1)that we can reach with only three terms is :

1/X+1/Y+1/(Z+1) . So:

6))........1/X+1/Y+1/(Z+1)+1/P=1

Subtract 6)) from 5))

7))........1/Z-1/(Z+1)=1/P

Now, all of the above is intended to be a proof that if we can find the smallest value for 1/Z, we can deduce the smallest value of 1/P and then of 1/E by substitution.

It seems obvious that if 1/X+1/Y+1/Z=1; the only solution within the constraints set in the given puzzle is: 1/2+1/3+1/6=1; and therefore the smallest possible value for 1/Z is 1/6. I hope this is obvious enough to be accepted as self evident, so I don't have to bother trying to prove it. So if Z=6, then substituting back through equations 7))....and 4)), we find P=42 and E=1806

Hmm
1/P-1/(P+1)=1/E

so:
((P+1) - P)/(P^2 + P) = 1/E

1/(P^2 + P) = 1/E

E = P^2 + P

That seems interesting...

works too, 1806 = 42^2 + 42

Aha, reusing t&b's cunning logic:


We want the smallest possible fraction for 1/E; such that:

1)).........1/A+1/B+1/C+1/D+1/E=1 ;
with A<B<C<D<E ..all integers.

To make 1/E as small as possible, we need the sum of the first four to be as large as possible. Suppose: that 1/P is the smallest possible fraction such that :

2))........1/M+1/N+1/O+1/P=1

If 1/P is as small as it can be, then the sum of the first three is as large as it can be. Clearly, then, the largest sum we can get (less than 1) with only four terms is :

1/M+1/N+1/O+1/(P+1)

So, to work out the smallest 1/E where 1/A + 1/B + 1/C + 1/D + /E = 1

We need to work out the smallest P such that 1M + 1/N + 1/0 + 1/P = 1
and then we add one to the denominator of P

Similarly, for the smallest P we will need the smallest S such that

1/Q + 1/R + 1/S = 1

and then add one to S

And for the smallest S we need the smallest U such that
1/T + 1/U = 1

and for the smallest U we need the smallest R such that 1/R = 1
i.e. R = 1

Remembering to add one to the denominator we can now derive that that T is 2

1/2 + 1/U = 1
This means U is 2, add one to denominator:

1/2 + 1/3 + 1/S = 1

this means that S is 6
1/2 + 1/3 + 1/7 + 1/P = 1

so P is 42, and

1/2 + 1/3 + 1/7 + 1/42 = 1

giving E = 1806

I think that is our proof isn't it? It was just a tiny adjustment of your logic.

Originally posted by iamatiger
Aha, reusing t&b's cunning logic:


We want the smallest possible fraction for 1/E; such that:

1)).........1/A+1/B+1/C+1/D+1/E=1 ;
with A<B<C<D<E ..all integers.

To make 1/E as small as possible, we need the sum of the first four to be as large as possible. Suppose: that 1/P is the smallest possible fraction such that :

2))........1/M+1/N+ ...[text shortened]... g E = 1806

I think that is our proof isn't it? It was just a tiny adjustment of your logic.

So the big computer in Hitchhikers guide was right! 42🙂