On Friday I posted this puzzle….

On Christmas Day my brother and I opened our presents, and then noticed a curious coincidence. If I were to give my bother seven of my presents, then I would have exactly as many presents as my bother.  On the other hand, if my brother were to give me seven presents, then I would have exactly twice as many presents as my brother. How many presents did each of us have?

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

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.

1. repton says:

Or: you both had infinitely many 🙂

2. I still like my way of doing it by replacing the number 7 with the number 1. So the answer is 7 and 5 which one can kind of do in ones head without stooping to algebra (which is kind of cheating). Then one multiplies 7 and 5 by 7 to get back to the original question.

3. I was too angry about how spoiled those kids were to concentrate on the maths for this one.

4. Barry Goddard says:

Yes too I deliberately did not solve this one as I could see at a glance that the ans must be a large number – certainly larger than 14 because

7 + 7 is fourteen

and

fourteen is seven times 2

I am even more amazed that the final ans is so much more larger than that. Richard and his bro must be very spoled indeed.

5. Maybe they were not that spoiled and they simply compared how many pieces of coal they got!

6. Eddie says:

Don’t see why it’s a coincidence…
Happy new year all. 2014. Bring it on.

7. Miss Chili says:

‘Oh, bother!’ said the editor. 😀

8. I did it the algebra way. At one point, where I got in a substitution tangle, Richard ended up with 7 (49) presents and his brother had -5 (-35), a situation that would truly cause bother for all.

9. If I (R) were to give my bother (B) seven of my presents, then I would have exactly as many presents as my bother:

R – 7 = B + 7
==> B = R – 14

if my brother (B) were to give me (R) seven presents, then I would have exactly twice as many presents as my brother:

R + 7 = 2(B – 7)

Replacing B with R-14:

R + 7 = 2 ((R -14) – 7) = 2(R – 21) = 2R – 42

subtract R on both sides:

7 = R – 42

==> R = 49

B = R – 14 = 35

Solution

Richard = 49
Brother = 35

10. Geoff Williams says:

R = Richard, B = Brother

R-7 = B+7
R = B+14

R+7 = 2(B-7)
R+7 = 2B-14
R = 2B-21

B+14 = 2B-21

B=35

R = 49

11. Lazy T says:

dohh, I did maths quickly in my head, got 35 and 21, decided you had been given a pack of cards, and didn’t bother to check properly, wottatwit.

12. Obviously it’s possible to solve this using simultaneous equations (as explained by others), but here’s how I did it fairly simply in my head:

Consider the state where both have equal numbers of presents. To get from here to the state where Richard has twice as many presents as his brother, his brother will pass 14 presents to Richard (the first 7 restores the initial state at the start of the puzzle, then 7 more to reach the ‘double’ state).

Passing 14 presents changes the difference in presents by 28 (subtracting 1 present from the brother makes a difference of 1; then adding that present on to Richard makes the difference be 2).
Since they were previously equal, Richard now has 28 more presents than his brother.

This means his brother must, at this point, also have 28 presents; then the 28 more is what makes Richard’s count double that, at 56.

So what were they to start with? That involves ‘undoing’ passing 7 from the brother to Richard; on other words, taking 7 from Richard’s total of 56 and adding 7 on to the brother’s 28. That gives the 49 and 28.

Each step, and the numbers involved, are simple enough to hold in your head.

13. Rena Vandiver says:

I love algebra too!

14. Bill T. says:

“algebra is cheating”, how is coming up with the corect answer using ANY valid method cheating?

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