CancelReport This Post

Please fill out the form below with your name, e-mail address and the reason(s) you wish to report this post.

 

Crossword Help Forum
Forum Rules

trevor

31st July 2009, 23:14
is it - 1?
a. b. c. d, are all 4?
141 of 1953  -   Report This Post

trevor

31st July 2009, 23:30
24 ? (mix of +'s .... oh, sugar . I give up.
142 of 1953  -   Report This Post

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.
143 of 1953  -   Report This Post

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
144 of 1953  -   Report This Post

trevor

1st August 2009, 01:15
Nope, not gettng it.
why assume "b=2 & c=2"?
145 of 1953  -   Report This Post

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
146 of 1953  -   Report This Post

celadon

1st August 2009, 09:22
If the sun lightens your hair, why does it darken your skin?
147 of 1953  -   Report This Post

moose

1st August 2009, 11:53
Why do homeless people always drink Tennants?
148 of 1953  -   Report This Post

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

149 of 1953  -   Report This Post

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
150 of 1953  -   Report This Post