is it - 1?

a. b. c. d, are all 4?

a. b. c. d, are all 4?

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24 ? (mix of +'s .... oh, sugar . I give up.

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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.

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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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

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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Nope, not gettng it.

why assume "b=2 & c=2"?

why assume "b=2 & c=2"?

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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

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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If the sun lightens your hair, why does it darken your skin?

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Why do homeless people always drink Tennants?

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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

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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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

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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