c4countdown

It requires no knowledge of maths beyond simple algebraic manipulation, but needs a systematic and intuitive approach:

Find all triples of positive integers a,b and c such that

(1 + 1/a)(1 + 1/b)(1 + 1/c) = 2

Good luck.

That looks like a STEP question...

(or maybe IMO, I suppose)

It's a BMO question, actually.

Conor wrote:It's a BMO question, actually.

BMO1 or BMO2? I sucked fairly hard at the former when I was at school (I got one right and a partial solution to another, which I think was near the median).

I've actually got a vague memory of seeing this one before, possibly got my wires crossed somewhere but remember it having a nice geometrical interpretation. I'll certainly have a stab at it tomorrow.

Paul Howe wrote:BMO1 or BMO2? I sucked fairly hard at the former when I was at school (I got one right and a partial solution to another, which I think was near the median).

I've actually got a vague memory of seeing this one before, possibly got my wires crossed somewhere but remember it having a nice geometrical interpretation. I'll certainly have a stab at it tomorrow.

BMO2, although it seemed a bit on the easier side for it. I'm pretty bad at Olympiad style questions, so it's always pleasing when I manage to do one. I'll have a look into a geometrical interpretation -- my method was purely algebraic.

Not sure if there's a faster way, but here goes...

It's a symmetrical identity so if (a,b,c) is a solution any permutation of these is also a solution. We can save time by finding the triples (a,b,c) with a<=b<=c and then permuting them to find all solutions.

If a = 1, then (1+1/b)(1+1/c) = 1, impossible for positive integer sols.

If a = 2, (1+1/b)(1+1/c) = 4/3. Trying b = 2 gives 1 + 1/c = 8/9, so that doesn't work.
b = 3 gives 1 + 1/c = 1, no positive integer solution
b = 4 gives 1 + 1/c = 16/15, so c = 15
b = 5 gives 1 + 1/c = 10/9, so c = 9
b = 6 gives 1 + 1/c = 8/7, so c = 7
b = 7 gives 1 + 1/c = 7/6, so c = 6, but here c < b which we're not considering, so can move on to a = 3.

a = 3 implies (1+1/b)(1+1/c) = 3/2
Working as before, b = 3 gives c = 8, b = 4 gives c = 5, and for b >=5 c is less than b, so move on to a = 4

a = 4 implies (1+1/b)(1+1/c) = 8/5, trying b = 4 gives c = 25/7 < 4, so no solutions for a = 4, and a similar situation pertains for all subsequent values of a.

So the answer is (2,4,15),(2,5,9),(2,6,7),(3,3,8),(3,4,5), and all permutations of these. I think the geometric interpretation was to multiply everything by abc so (a+1)(b+1)(c+1) = 2abc. The solutions then describe the set of all cuboids having integer length sides such that increasing the length of each side by 1 doubles the volume. Cute, but not really relevant to solving the problem.

Paul Howe wrote: So the answer is (2,4,15),(2,5,9),(2,6,7),(3,3,8),(3,4,5), and all permutations of these. I think the geometric interpretation was to multiply everything by abc so (a+1)(b+1)(c+1) = 2abc. The solutions then describe the set of all cuboids having integer length sides such that increasing the length of each side by 1 doubles the volume. Cute, but not really relevant to solving the problem.

I did it this far and, having spotted the geometrical significance, couldn't think of anyway to use it to solve the problem (since geometry isn't constrained to integers.) Did I miss something, or was the geometrical bit just for "fun"?

Charlie Reams wrote:
I did it this far and, having spotted the geometrical significance, couldn't think of anyway to use it to solve the problem (since geometry isn't constrained to integers.) Did I miss something, or was the geometrical bit just for "fun"?

No, not relevant, I saw this in a solution booklet when I was practising for BMO yonks ago and had a fuzzy memory of it having geometrical flavour, but couldn't think of the significance when I posted last night. It would've been a nicer question if they'd given the geometry description and required you to translate into algebra, I think.

Paul gets it, all permutations, well done. Nice alternative method to mine, I can see it being useful. The geometrical interpretation was interesting too.

What I did: Multiply through by abc to get (a+1)(b+1)(c+1) = 2abc. Let a = 2, so 3(b+1)(c+1) = 4bc.
Rearrange to get bc - 3b - 3c - 3 = 0 and "factorize" to end up with (b - 3)(c - 3) = 12, and you can consider factors of 12 -- 1*12, 2*6 and 3*4. And like your method, proceed for other values of a.

I don't know how you guys do all that stuff - I got brain ache just reading the question!!

I took a completely different tact. I tried to find improper fractions for (1 + 1/a), (1+1/b) and (1+1/c) like d/e, e/f, f/g such that when you times them together some of the numerators and denominators cancel and you're left with d/g where d is twice g. Through trial and error on the bus home in my head I got all of Paul's solutions except 3,3,8, so I'm quite pleased with that

Also I know you said integers but there's also: 2,3,infinity and 1,infinity,infinity

Dinos Sfyris wrote:Also I know you said integers but there's also: 2,3,infinity and 1,infinity,infinity

But 1/infinity isn't defined...

Michael Wallace wrote:

Dinos Sfyris wrote:Also I know you said integers but there's also: 2,3,infinity and 1,infinity,infinity

But 1/infinity isn't defined...

But if it were it would be zero, yes? :p

Dinos Sfyris wrote: But if it were it would be zero, yes? :p

You can say as a tends to infinity, 1/a tends to zero. But 1/infinity isn't defined since infinity isn't a number.

And in any case the question asked for integers, and infinity definitely ain't one of those.

Charlie Reams wrote:And in any case the question asked for integers, and infinity definitely ain't one of those.

To be fair, he did acknowledge that bit. I just get unreasonably annoyed when people say things like 1/0 = infinity

Michael Wallace wrote:

Charlie Reams wrote:And in any case the question asked for integers, and infinity definitely ain't one of those.

I just get unreasonably annoyed when people say things like 1/0 = infinity

I just went to a very weird visual place involving a raccoon-like Hulk!