 ## Friday, June 15, 2012

### A Dinner Puzzle

Recently at a dinner party the hostess offered the following puzzle for anyone interested. Assume you have two numbers: a and b. Create a math sentence using these numbers and the +,-,/,*, square root and power signs so that this sentence will produce the largest of these numbers as the output.

The hostess told that this puzzle was offered to her a day before and she couldn't concentrate for a number of hours till she solved it. Needless to say, I also do not remember what was served at this dinner or the names of the guests. It took a whole evening and finally at 11:30pm the angel of solution descended upon me. Can you catch him as well?

Image by Amelia-Jane, distributed under CCL.

Your answers are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.

#### 12 comments:

John Golden said...

This seems ambiguous. Assumptions: no info about the numbers, assume a and b are real, you can use each only once, root and power sign mean x^a or x^(1/a)... just unsure what the goal is. Probably I'm being too mathy, but it's hard to imagine an answer that would be good for all a and b. Even if they're counting numbers, maybe. Unless it's like an alphabetical answer? I look forward to other people's solutions this week.

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Ryan said...

assuming a>b

a+ (a/b) * ( rt( (a+b)^2 ) ) - (a/b) * ( rt( (a+b)^2 ) )

Were we supposed to use each operator/variable once? I'm also assuming we can't use other numbers than a, b, and the 2 as a power/root, since the use of 0 would make things too easy. If we can't use 2 we could replace it with "a" in the root/power.

Jerome said...

I don't know if the problem is to choose the largest number (a or b) and then use an algorithm that will leave one of them using +-*/sqrt and power, or if the problem is to determine the largest number itself and let it pop out of the fire.

If it is the first, I suppose you could do it like this (although there should be many solutions)

sqrt( [a + b]^2 - 2*a*b - (a^2 + b^2))^n + ab
==============================================
b

where a is the largest number.

The other problem escaptes me right now.

Anonymous said...

Your puzzle does not indicate if you can use numbers or other letters indicating quantities. Assuming that a and b are both positive quantities, I would venture that ab to the ab power is the highest number.

Gurubandhu

Maria said...

To clarify a bit. All you need to do is come up with a sentence of a type: a+b-a*b-sqrt(a...b)..... that will produce as an output the largest of two numbers, a and b. If a>b then the output should be a, if a<b then the output should be b.
The answer is simple and elegant. It is easier to get to it if you draw two rectangles of different length, representing a and b, next to each other. How to manipulate them to find the largest?

if-then statements are not allowed in this sentence, only +,-,*,/,sqrt and ^2 signs.

good luck!

Jerome said...

I did this with the quadratic equation. There are numerous cases to consider. I will only consider the one: a>0 and b<0 The others would be done much the same way. In addition one or both a or b could be zero. Those two will be ignored.

let one of the roots of the quadratic be a and the other be - b

Therefore the quadratic looks like this.

x = [-(-a + b) +/- SQRT( (-(a - b)^2 + 4a*b))/2

We are only concerned about the right hand side.
[a - b + sqrt((a - b)^2 + 4ab)] / 2

[ a - b + sqrt(a + b)^2] / 2
[ a - b + (a + b)/2

2a / 2 = a

The larger number comes out as an a.

Jerome said...

I need to note that you could set this up with another variable such that you would not be working with negatives.

Ilya said...

I came up with an answer in a somewhat abstract way before I read your suggestion of a geometrical approach. What I found helpful is using the fact the combination of square and square root yields an absolute value of a number. With that, (a-b)+SQRT((a-b)*(a-b)) will be equal 0 if a <=b or (a-b)*2 if a > b. If we multiply that by 2 and then add b, we get exactly what is needed:
((a-b)+SQRT((a-b)*(a-b)))*2 + b, which, when a <= b, will be equal b, and when a > b, will be equal (a-b)+b = a.

Maria said...

OK, I think the puzzle may be a bit too engineering and abstract that our usual puzzles. But some of you have got it right.

Imagine two rectangles, one bigger than another. Their sum + their difference, together divided by 2 will give the length of the largest rectangle.
So, ((a+b) + |a-b|)/2 = largest of a and b

However, we do not have a module || operation, so the trick is to replace it with sqrt ((a+b)^2).

Therefore, the answer is:

((a+b) + sqrt((a+b)^2))/2

Does it make sense now?
A puzzle point for Ilya, Jerome and Ryan who produced close answers.

Anonymous said...

looking at your answer, it seems to me as though
((a+b) + sqrt((a+b)^2))/2
should equal a+b -- not the largest of a and b

sqrt((a+b)^2)) = sqrt((a+b)(a+b)) = a+b

((a+b) + (a+b))/2 = 2*(a+b)/2 = a+b

Am I missing something?

Thanks,
TracyZ

Maria said...

My mistake - it should be "a-b" instead of "a+b".

We are replacing a module || operation with sqrt ((a-b)^2).

Therefore, the answer is:

((a+b) + sqrt((a-b)^2))/2

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