Math Is Fun Forum

Hi Bobby,

Hi mathsyperson,

Sneaky or not - I like it! Well spotted!

This puzzle is at least 40 years old, and no one was that sneaky back then, especially not any cross-number puzzle setters - excluding Rhombus, of course!

Yes - "the puzzle is solvable without this style of thinking". I know that because I solved it, and I'm not sneaky enough to have come up with that trick.

But I just wish I'd thought of it first!

Hi Bobby,

Now there's a question I've never considered before about any cross-number puzzles, or been asked!

Sorry, Bobby, but I don't think I should give the answer to your question because it would colour your thinking - and hence your solving strategy, unnecessarily. It would allow you to make certain assumptions and that might spoil the true logic path. All the information needed to solve the puzzle is there.

Finding a starting point should take a fair bit of good sleuthing work, and even then the rest of the puzzle certainly won't just fall into place. It's a good test of logic skills right to the end.

Hi Bobby,

Yes, that's right.

And if, for example, you then solve 18 across to be 321, the 3 would go in r7c5 and the 2 in r7c6. The 1 in r7c7 is already there from when you entered it as the second digit of 16 down.

Same idea as in a standard crossword puzzle.

Hi Bobby,

Likewise. I did it on paper (in the same way as in my image) by drawing a grid with R,RD,RS,B,BD,BS,G,GD,GS,Y,YD,YS,A,AD,AS,W ("D" as in "dot" for "spotted" and "A" for grey) down the left-hand side and numbers 1-16 across the top, and dividing those numbers into four groups (as in the image) to represent the grid rows.

I then solved it in the manner typical of this kind of logic puzzle...by putting a cross in the grid boxes where a coloured square couldn't possibly go.

Then, by continually scanning the clues, more crosses are added, to the point where only the solution remains. After solving a clue I crossed out the clue number so I knew not to look at it again...and on the image that is represented by the green squares to the left of the clues.

The image was just my attempt at giving an easy-to-understand way of explaining my strategy. It shows how it can be done neatly (I scribble!) - and on Excel you can also backtrack if you blunder.

The solution is unique, which is confirmed by my strategy.

Hi, Bobby,

Took me about an hour to an hour and a half, I think - from working out a strategy to solving it.  I then spent at least that long preparing a graphical solution in case anybody ever asked me about it.

Careful and accurate progress is the key with this type of puzzle - because it's quite long and there are so many statements - and, unlike with some other puzzles, it's hard to backtrack once you discover you've muffed it.

It's so easy to try to speed along, but just as easily you can overlook the tiniest little thing and then you have to start again...which ends up taking so much extra time in the long run.

As with a building project: "Measure twice, cut once".

Hi knightdrak01,

Nah...I'm not even in the same ball park as bobbym and soroban! I just have a go at some of the easier stuff that doesn't require advanced maths knowledge.

No doubt the setter of your two problems has a juicy eye-opening solution for each...ones that I'm struggling to find. I'm often amazed at the clever strategies used by others in their solutions to tricky brain-teasers!

That's one of the reasons maths is fun!

And post #4...and the thread title.

The OED's definition of reasoning: To think in a connected, sensible, or logical manner; to employ the faculty of reason in forming conclusions.

I can't do any of that, so I'm not going to be of much help here!

Hi Bobby and knightdrak01,

Bobby: You added the 10-sweets shortfall instead deducting it.

Here's how I went about it...
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Trial and error:

I guessed 10 kids to start with, so...
1. (10 kids x 8 sweets) + 26 sweets = 106 sweets
2. (10 kids x 12 sweets) - 10 sweets = 110 sweets

It's obvious that to end up with the same number of sweets for scenarios 1 & 2 the number of kids has to be reduced from 10, so...
1. (9 kids x 8 sweets) + 26 sweets = 98 sweets
2. (9 kids x 12 sweets) - 10 sweets = 98 sweets

Solved (same number of sweets for each scenario): there are 9 kids.
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Reasoning strategy (which really is algebra in words, though):

There are two constants:
1. The number of kids.
2. The total number of sweets the kids receive...which is the same for each scenario.

The number of sweets given to each kid in each scenario and the number of sweets over or under Uncle Ben's supply are the variables.

Those variables:
1. The difference between the excess and shortfall quantities is 36: ie, 26 - (-10).
2. The difference between the number of sweets each kid gets is 4: ie, 12 - 8.

Divide the total quantity difference (36) by the difference between the number of sweets each kid gets (4) to find the number of kids: 36 / 4 = 9 kids.
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Algebra:

"k" = kids
8k + 26 = 12k - 10
4k = 36
k = 9
So again, there are 9 kids.

Hi Denominator,

Actually the backslash (\) should be a forward slash (/)

The 'hide' feature, and some others, are covered here.

You can also find out how someone did something or other just by clicking the "Quote" tab in the bottom right-hand corner of their post...but you must be logged in to see that tab.

I think your avatar does it for you, Bobby (I don't know what else would).

Hi! Here's some figuring that stem from the inconsistencies in the given data:

ACD's area (32m²) yields approx 11.3137 metres for AD, which differs from the text's 11.31 metres and the diagram's 11.3 metres.

There are therefore 3 different answers, and, using the sine rule as per simplyjasper's post:
1. If ACD's area is 32m²: BD = 2 x √32 x sin 79 / sin 56 = approx 13.396 metres
2. If AD is 11.31 metres (text): BD = 11.31 x sin 79 / sin 56 = approx 13.391677 metres
3. If AD is 11.3 metres (diagram): BD = 11.3 x sin 79 / sin 56 = approx 13.3798 metres

I suspect that AD's given length is just an approximation.