solving this

**runrunrun** · December 4th 2006, 12:48 PM

my first puzzle,,,,let's play with puzzles

The figure below is a number pyramid. The number in each square is the sum of the numbers in the two squares on which it rests (except for the squares in the bottom row). As you can see some of the numbers are hidden.
Determine the exact value of the hidden number in the square labeled with the question mark. It needn't be a whole number.

good luck

**topsquark** · December 4th 2006, 01:04 PM

Originally Posted by runrunrun

my first puzzle,,,,let's play with puzzles

The figure below is a number pyramid. The number in each square is the sum of the numbers in the two squares on which it rests (except for the squares in the bottom row). As you can see some of the numbers are hidden.
Determine the exact value of the hidden number in the square labeled with the question mark. It needn't be a whole number.

good luck

(It had better not be a whole number!)
Whoopsies! I gave my "x" not the answer.
I got 71/3.

-Dan

**CaptainBlack** · December 4th 2006, 01:07 PM

Originally Posted by runrunrun

my first puzzle,,,,let's play with puzzles

The figure below is a number pyramid. The number in each square is the sum of the numbers in the two squares on which it rests (except for the squares in the bottom row). As you can see some of the numbers are hidden.
Determine the exact value of the hidden number in the square labeled with the question mark. It needn't be a whole number.

good luck

select space below if you want to see my answer

23 2/3

Write D in the centre box on the bottom, then fill in all the empty cells in terms of D and what is
given. Then the relationship between the second from top row and the top row gives an equation
from which D can be found. Having found D all the cells may now be completed including the one
marked with a ?.

RonL

**TriKri** · December 4th 2006, 01:25 PM

Select the text below to show my answer.

If we turn it all upside down, it's like almost like Pascal's triangle. If all the numbers in the bottom row except from the one would be zero, for example, the number in the top square would be [tex]5\choose{1}[/tex]. But since we now have five different bottom row squares, of which one is unknown (let's call that number a), which all affect the top square, the number in the top square is now a linear combination of five different combination quantities (if I may call it so). The equation for the number in the top square would be:

[tex]81 = 1\cdot {5\choose{1}} + 2\cdot {5\choose{2}} + a\cdot {5\choose{3}} + 3\cdot {5\choose{4}} + 4\cdot {5\choose{5}}[/tex]

Then we have the equation for the square with the question mark in it, and that equation looks like this:

[tex]? = 2\cdot {3\choose{1}} + a\cdot {3\choose{2}} + 3\cdot {3\choose{3}}[/tex]

**runrunrun** · December 5th 2006, 03:18 PM

so many answer, which one is correct. huh ? what the numbers in the box.

**TriKri** · December 6th 2006, 04:30 AM

It seems like my previous post where incorect, since I once again forgot to use zero indexing inside the parenthesis. So, it really should look like this:

If we turn it all upside down, it's like almost like Pascal's triangle. If all the numbers in the bottom row except from the one would be zero, for example, the number in the top square would be $4\choose{0}$ . But since we now have five different bottom row squares, of which one is unknown (let's call that number a), which all affect the top square, the number in the top square is now a linear combination of five different combination quantities (if I may call it so). The equation for the number in the top square would be:

$81 = 1\cdot {4\choose{0}} + 2\cdot {4\choose{1}} + a\cdot {4\choose{2}} + 3\cdot {4\choose{3}} + 4\cdot {4\choose{4}}$

Then we have the equation for the square with the question mark in it, and that equation looks like this:

$? = 2\cdot {2\choose{0}} + a\cdot {2\choose{1}} + 3\cdot {2\choose{2}}$

So, first solve what $a$ is, then you solve what $?$ is ( $?$ depends on $a$ ) to get what should be in the square. This I hope will be right, but I still haven't tested it. $?$ contains the answer.

$81 = 1\cdot {4\choose{0}} + 2\cdot {4\choose{1}} + a\cdot {4\choose{2}} + 3\cdot {4\choose{3}} + 4\cdot {4\choose{4}}$

$81\ =\ 1\cdot 1\ +\ 2\cdot 4\ +\ a\cdot 6\ +\ 3\cdot 4\ +\ 4\cdot 1$

$81-1-8-12-4\ =\ 6\cdot a$

$a\ =\ \frac{56}{6}$

$? = 2\cdot {2\choose{0}} + \frac{56}{6}\cdot {2\choose{1}} + 3\cdot {2\choose{2}}$

$? = 2\cdot 1 + \frac{56}{6}\cdot 2 + 3\cdot 1$

$? = 2 + \frac{56}{3} + 3$

$? = 23\ 2/3$

**CaptainBlack** · December 6th 2006, 05:00 AM

Originally Posted by runrunrun

so many answer, which one is correct. huh ? what the numbers in the box.

Selfevidently it is mine

RonL

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