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trevor

31st July 2009, 23:14
is it - 1?
a. b. c. d, are all 4?
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trevor

31st July 2009, 23:30
24 ? (mix of +'s .... oh, sugar . I give up.
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john (from arran)

31st July 2009, 23:37
No, it's more than 1.

Anyway it's time for my cocoa now so I'll leave you all to think about it and give you the answer tomorrow.
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jimc

1st August 2009, 00:57
There are an infinite number of solutions.

If we take b=2 & c=2 then, for example, one solution has a=-3 & d=3:

  1/(-3) + 1/2 + 1/2 + 1/3 = 1

Another solution would be with a=-4 & d=4:

  1/(-4) + 1/2 + 1/2 + 1/4 = 1

Any (of the infinite number of integer) values for a & d that cancel each other out would work.

JimC
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trevor

1st August 2009, 01:15
Nope, not gettng it.
why assume "b=2 & c=2"?
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jimc

1st August 2009, 08:35
Hi Trevor.

Sorry for the delayed reply - sleep got me.

Taking b=2 & c=2 just let me show easily why there might be an infinity of solutions.

There are other solutions in which b & c are not = 2 (such as the one you pointed out in which a=b=c=d=4) but the conclusion (that there are an infinity of solutions) still holds.

JimC
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celadon

1st August 2009, 09:22
If the sun lightens your hair, why does it darken your skin?
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moose

1st August 2009, 11:53
Why do homeless people always drink Tennants?
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john (from arran)

1st August 2009, 11:54
Ah, so JimC has introduced negative numbers. I guess since I didn't say they all had to be positive that's good lateral thinking. Here's the solution I worked out when this puzzle was posed to me a few weeks ago, giving 14 solutions in total:

a != 0.

If a = 1, then 1/b = 1/c = 1/d = 0, which cannot be true for
positive finite integers, so a > 1.

If a >= 5, then b, c, and d are all >= 5, and the sum
of (1/a + 1/b + 1/c + 1/d) is at most 4/5 < 1, so a < 5.

Therefore a = 2, 3, or 4.

Case A2. Let a = 2.
2 <= b <= c <= d, and 1/b + 1/c + 1/d = 1/2.
If b = 2, then 1/c = 1/d = 0, so b != 2.
If b > 6, then c and d are both >= 7, and the sum
of (1/b + 1/c + 1/d) is at most 3/7 < 1/2, so b <= 6.
Therefore, b = 3, 4, 5, or 6.
Case A2B3. Let b = 3.
3 <= c <= d, and 1/c + 1/d = 1/6.
Using reasoning similar to the above,
we have 6 < c <= 12.
Therefore, c = 7, 8, 9, 10, 11, or 12.
Rearranging the reciprocal sum and solving for d,
we have d = 6c / (c - 6).
Plugging in the possibilities for c, we find
the following solutions (none for c = 11):
c = 7, d = 42
c = 8, d = 24
c = 9, d = 18
c = 10, d = 15
c = 12, d = 12
Case A2B4. Let b = 4.
4 <= c <= d, and 1/c + 1/d = 1/4.
Reasoning as above, 4 < c <= 8.
Therefore, c = 5, 6, 7, or 8.
d = 4c / (c - 4).
Solutions are (none for c = 7):
c = 5, d = 20
c = 6, d = 12
c = 8, d = 8
Case A2B5. Let b = 5.
5 <= c <= d, and 1/c + 1/d = 3/10.
Reasoning as above, 5 <= c < 7.
Therefore, c = 5 or 6.
d = 10c / (3c - 10).
Solutions are (none for c = 6):
c = 5, d = 10.
Case A2B6. Let b = 6.
6 <= c <= d, and 1/c + 1/d = 1/3.
Reasoning as above, 6 <= c <= 6.
Therefore, c = 6.
d = 3c / (c - 3).
Solutions are:
c = 6, d = 6

Case A3. Let a = 3.
3 <= b <= c <= d, and 1/b + 1/c + 1/d = 2/3.
Reasoning as above, 3 <= b < 5.
Therefore, b = 3 or 4.
Case A3B3. Let b = 3.
3 <= c <= d, and 1/c + 1/d = 1/3.
Reasoning as above, 4 <= c <= 6.
Therefore, c = 4, 5, or 6.
d = 3c / (c - 3).
Solutions are (none for c = 5):
c = 4, d = 12
c = 6, d = 6
Case A3B4. Let b = 4.
4 <= c <= d, and 1/c + 1/d = 5/12.
Reasoning as above, 4 <= c < 5.
Therefore, c = 4.
d = 12c / (5c - 12).
Solutions are:
c = 4, d = 6

Case A4. Let a = 4.
4 <= b <= c <= d, and 1/b + 1/c + 1/d = 3/4.
Reasoning as above, 4 <= b < 5.
Therefore, b = 4.
Case A4B4. Let b = 4.
Solutions are:
c = 4, d = 4

Summary of solutions:
a = 2, b = 3, c = 7, d = 42
a = 2, b = 3, c = 8, d = 24
a = 2, b = 3, c = 9, d = 18
a = 2, b = 3, c = 10, d = 15
a = 2, b = 3, c = 12, d = 12
a = 2, b = 4, c = 5, d = 20
a = 2, b = 4, c = 6, d = 12
a = 2, b = 4, c = 8, d = 8
a = 2, b = 5, c = 5, d = 10
a = 2, b = 6, c = 6, d = 6
a = 3, b = 3, c = 4, d = 12
a = 3, b = 3, c = 6, d = 6
a = 3, b = 4, c = 4, d = 6
a = 4, b = 4, c = 4, d = 4

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jimc

1st August 2009, 12:25
Hi John o'A.

I think that's a really clever thing that you did there. Did the wine help before or after you had solved it (or both)?

Arran must be a peaceful place if you are able to get that worked out without interruption!

Sorry about the negative integers - I thought that, maybe, this was one of those puzzles that didn't need the brute force of working out each and every solution, but just some 'trick' or other.

JimC
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