On Friday I posted this puzzle….

Imagine that there are 2 trees in a garden. Let us call them Tree A and Tree B. Now imagine that there some birds on both trees.

The birds on Tree A say to the birds on Tree B: ‘If one of you comes to our tree, then our population will be the double of yours.’

Then birds on Tree B say to the birds on Tree A: ‘If one of you comes to our tree, then our population will be equal to that of yours.’

How many birds are there in each tree?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

There are 7 birds on Tree A, and 5 on Tree B. Did you solve it?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle**(UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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Tried and failed to solve this with algebra. Must be rusty. But I solved it thus:

Assume each tree contains whole birds.

If A+1 is *double* B-1 then A must be an even number.

If B+1 is the same as A-1 then A must be B+2.

B must be an odd number that is two less than A.

Given that A+1 = 2(B-1) and B+1=A-1 then A and B must be small numbers.

Therefore, candidates are 3,1;5,3; 7,5; 9,7.

It is trivial to determine which of these four pairs is the correct answer

A=7; B=5.

I know what you mean by rusty, lol, but to shake off the rust, you just have to read slowly and rewrite what the problem says using letters(variables) This gives you TreeA=X TreeB=Y …. Taking 1 from B and adding it to A, makes treeA twice that of treeB so… X+1=2(Y-1) And taking 1 from treeA and giving it to B makes them equal so …. X-1=Y+1 So simplifying gives X-2Y+3=0 and X-Y-2=0 and you just have to solve these simultaneous equations! Cheers!

I got 3 birds on each tree :

“The birds on Tree A say to the birds on Tree B: ‘If one of you comes to our tree, then our population will be the double of yours.’”

3 – 3 >>> 4 – 2

“THEN birds on Tree B say to the birds on Tree A: ‘If one of you comes to our tree, then our population will be equal to that of yours.’

4 – 2 >>> 3 – 3

I thought the word “THEN” would be important to solve this puzzle…

Ahem. *”then A must be an odd number”.

I counted on my fingers like a boss.

Hmm i got 6 on Tree A and 4 on Tree B. Does this work?

The question gives the following equalities:

(A+1) = 2(B-1)

(A-1) = (B+1)

your numbers give (7) = 2 * (3), though they satisfy the second equality

Anon- I got 6 and 4 with algebra. It was an error with the brackets. It’s along the lines of 2(b-1) which = 2b-2 not 2b-1 which is where I went wrong.

I got the answer in the end by working out that the answer had to add up to an even number and be divisible by 3 . Then I got my penny jar out to work out the combinations .

I just solved two very simple simultaneous equations.

Simple Algebra:

Statement 1 gives:

A + 1 = 2 x (B – 1)

A = 2B – 3

Statement 2 gives:

A – 1 = B + 1

A = B + 2

So:

2B – 3 = B + 2

B = 5

Then A = 5 + 2 = 7

We can count, you know.

This is what I did, but it’s been so long since I took algebra, is that the “simultaneous equations” approach people are talking about? I remember learning to do simultaneous equations, but nothing concrete about how. I just figured if I could isolate one of the variables in both equations, then I could set the other sides equal to each other and solve for one variable. Did I remember the correct approach, or just happen to make up something that also worked?

Eddie – The issue isn’t whether you can count. It’s whether you can think.

There’s not much to think about in this puzzle. You could also use the matrix approach, finding the inverse of the 2 by 2 matrix.

I found a 2nd solution, (3 on each tree) because of my interpretation of the word “then”.

“Then birds on Tree B say to the birds on Tree A: ‘If one of you comes to our tree, then our population will be equal to that of yours.’ ”

I presumed you were being tricky and I interpreted the word “then” as meaning AFTER one of the B birds flew to A.

Stan

That doesn’t work though, because the words spoken by the birds start “If”; the numbers are all hypothetical: nowhere does it say that any of the birds actually fly between the trees.

The “then”, however, is said by the narrator, not by any of the birds. So the “then” doesn’t feature in the hypothetical bird-moving. If the second set of birds to speak had started their statement with “then”, that would’ve been the birds saying their statement follows on from the situation established by the first birds.

But because the then is merely describing the order in which the birds speak, both statements made are hypothetical alternatives based on the current situation.

Yep I thought the same as Stan. 3 in each tree. For the same reasons.

The quick and easy way is thus. Clearly answers must be fairly small numbers. Since in first move, there are twice as many in one tree as in other, the total number must be divisible by 3. Since in second move there are equal birds in each tree, total must be even. So number of birds is divisible by 6. 6 clearly isn’t enough. Try 12, oh yes that words.

Yes got it, it also works for two copses in a field where the trees of copse A say to those of copse B “If one of you….”

It doesn’t require algebra, nor trying out (and eliminating) a few possibilities.

1 bird changing trees changes the difference between the trees’ populations by 2.

In order for the second statement (‘equal’) to be true, the trees initial populations’ must therefore differ by 2.

So 1 bird flying in the opposite direction, as for the first (‘double’) statement must therefore yield a difference of 4.

When when population is double another and their difference is 4 then obviously the smaller population also has to be 4, and so the larger population is 8.

Which means that before that bird changed, the original populations were 5 and 7.

s/When when/When one/

Your solution is still algebra. You’re just doing it in a more “wordy” manner.

While its true that this is such a simple problem that using algebra may not be necessary, its still a good idea to understand how to do it using algebraic equations because if the problem was even slightly more complicated you would not be able to “reason” it out. Knowing how to extract the equations becomes integral to solving harder problems, so its good practice.

One Eyed Jack: Sort-of. And I considered that too. But my solution isn’t a direct transliteration of how I’d solve the simultaneous equations, nor of any the other algebraic answers given by others.

And I think my explanation is simple enough that it could be followed by somebody who struggles with algebra.

Whereas translating the logic I followed into algebra makes it look much more complicated:

|(|A – B| – |(A + 1) – (B – 1)|)| = 2

A + 1 = B – 1

|A – B| = 2

|(|A – B| – |(A – 1) – (B + 1)|)| = 2

|(2 – |(A – 1) – (B + 1)|)| = 2

|(A – 1) – (B + 1)| = 0 or 4; |(A – 1) – (B + 1)| > |A – B|, so |(A – 1) – (B + 1)| = 4

(B + 1) > (A – 1), so (B + 1) – (A – 1) = 4

B + 1 = 4 + (A – 1)

(B + 1) = 2(A – 1)

0 = -4 + A – 1

5 = A

I think that’s far harder to follow than the other algebraic solutions given (and so doesn’t have much merit as an approach).

Whereas my ‘wordy’ solution has a simplicity and elegance. So I don’t think it’s merely algebra but using words instead of symbols.

Got it.

That’s not a puzzle, that’s a maths problem! I thought the answer must be along the lines of ‘We don’t know because the birds might be lying’

So long as the birds in question aren’t mynah birds, there oughtn’t be much of a problem. Whichever bird species they are, they do seem to be smarter than the average…bear? * smirk *

Again, there is no “puzzle” here, especially for the intelligent people who read this blog, its just straight algebra! I guess though it is good to practice your algebra skills, you never know when it may come in handy. The equations were: X-2Y+3=0 and X-Y-2=0 giving X=7(treeA) and Y=5(treeB)

I see the puzzle as parsing the question to form the simultaneous equations.

Good! I solved the puzzle AND I don’t understand anything of the solutions given above.

Hurrah! \o/

Interesting mathematical solutions but I’m afraid they’re all wrong. There is no real answer to the problem as it is clearly a trick question – birds can’t talk!

Three birds in each tree, for six total.

One bird goes from Tree B to Tree A leaving four in Tree A and two in Tree B. Then one bird goes to Tree B from Tree A leaving the same in each tree.

They didn’t actually move.

The Birds can Talk ?

6

7 and 5

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