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Who can solve this puzzle?

You have to find the missing numerical value in the equations below.

The small symbols are integers and the big symbols (with small symbols inside them) stand for functions or series. ##### Share on other sites

Maybe it's a little too hard. Let me help you on the way a bit.

Example:

From the solutions to the upper left-most equations, you could deduce that symbols standing next to each other have to be added, since 19 is a prime number.

Then if you choose the following values for the small square, triangle and diamond, square = 1, triangle = 10, diamond = 6, the upper left-most equation is solved.

Moving on to the equation below that one. If you choose the Fibonacci function for the big circles, you get Fib(6) + 10 + Fib(1) = 19, which is correct as well, since Fib(6) yields 8 and Fib(1) = 1.

Of course this isn't the correct solution, I'm just showing you how it might be done.

The big symbols could stand for all kinds of functions, e.g. squares, cubes, square roots, prime numbers, triangular numbers, factorials, etc.

Koko

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Nobody seems to have been able to solve this puzzle yet. Maybe I should give another clue.

It is the most efficient to start with the equations with the numerical values:

s+s+s+t+d = 19, C(s)+C(d)+c = 21, C(d)+t+C(s) = 19.

c = cirlce, s = square, t = triangle, d = diamond.

The small symbols all stand for distinct non-negative integers.

Now, if you can find out which of the previously mentioned functions the C() stands for, you might also be able to find out what the S(), T() and D() stand for and solve the final equation.

Koko

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is this college math?

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s+s+s+t+d = 19, C(s)+C(d)+c = 21, C(d)+t+C(s) = 19.

Shouldn't that be C(s)+C(t)+c = 21 ?

Is there definitely enough information there to solve everything? When symbols are juxtaposed does that always mean addition?

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Shouldn't that be C(s)+C(t)+c = 21 ?

Is there definitely enough information there to solve everything? When symbols are juxtaposed does that always mean addition?

You're absolutely right, it should be C(s)+C(t)+c = 21. And yes, symbols standing next to each other always have to be added, no subtracting.

There is enough information available, you just have to work it out step by step.

When you try each of the following functions for C() there aren't gonna be a whole lot of possibilities for which you can solve the three equations. (s+s+s+t+d = 19, C(s)+C(t)+c = 21, C(d)+t+C(s) = 19)

C() could be Square(), Cube(), SquareRoot(), Fibonacci() or TriangularNumber().

You may have to use other functions for S(), D() and T(), but the function for C() is in this list above.

When you have found a set of solutions for these three equations you can then move on to to the next equations.

Obviously, there is only one solution possible!

Good luck!

Koko

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only the 1st 2 make any sense to me. I just add up the sides to each, and in the second one I do the same and take away the amount of sides in the big shape its inside of.

as for the rest of the stuff You mention I`v never even heard of some of them!

and I despise maths anyway.

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Squares: 1 4 9 16 25 36 49 ... etc

Cubes: 1 8 27 64 125 216 343 ... etc

Fibonacci: 1 1 2 3 5 8 13 ... etc

Triangular Numbers: 1 3 6 10 15 21 28 ... etc

So, if s = 7, d = 2 and t = 5 and C() would be Fibonacci(), then the following equation could be made C(d) + t + C(s) = 19, because Fibonacci(2) = 1, t = 5 and Fibonacci(7) = 13, 1 + 5 + 13 = 19.

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Here's another clue:

The big circles stand for triangular numbers. So a big triangle with number 9 inside would stand for the 9th triangular number, which is 45. (1-3-6-10-15-21-28-36-45)

Now all you have to do is find out what the other symbols stand for.

The small symbols are all numbers in the range 0-10.

Koko

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I am not sure about T( ), it’s a little strange function. But on the way I have guessed numbers and functions could be, it works:

[hide] c=0; r=1; t=3; s=5

C( )= triangular numbers; S()= cube

I don’t know if D( ) has a name, but it could be

D(1)=1; D(2)= 1x2=2; D(3)= 1x2x3=6; D(4)= 1x2x3x4= 24 ; D(5)= 1x2x3x4x5=120….

My guess for T( ) would be:

Addition of all the numbers that compose the whole figure raise to the power of the number of figures that are in the whole figure:

T (1)= 1 to the power of one=1

T (10)= 1 to the power of two=1

T (15)= 6 to the power of two= 36

T(720)= 9 to the power of three= 27

With all this data, the solution would be ?=200[/hide]

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Hey zule,

Nice job! You're almost there. All the assumptions you made about the small symbols and the functions D() and S() are correct.

Only for T() you haven't found the correct function, btw 9 to the third power is 729.

Koko

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It always happens to me: Instead failing in the difficult taxes, I make the most stupid mistakes:doh:. So, I think the solution could be:

[hide]T( ): put the number the other way around.

T(10)=01

T(5)=5

T(6)=6

T(720)=027

So, the solution will be ?=101

Now, I am going to go over the calculation to test if I have made another mistake.[/hide]

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Yup! That's it! Well found!

Koko

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