3 Questions

[matt8441]

I need some help on how to do these:

How many (unordered) triplets of real numbers {x,y,z} are there such that
x(x+y+z)=26; y(x+y+z)=27; z(x+y+z)=28?

A die consists of a cube which has a different color on each of 6 faces. How many distinguishably different kinds of dice can be made?

What is the sum of the positive integers less than ten million having exactly 77 divisors?

Would the last one just be 0? Because 77! is far over ten million.

 

[pka]

For the first problem here are some possibilities.
Multiplying out and adding we get \(\displaystyle \L
\begin{array}{l}
x^2 + xy + xz = 26 \\
y^2 + xy + yz = 27 \\
z^2 + xz + yz = 28 \\
x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 81 \\
\end{array}\).

Also note \(\displaystyle \L
x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = (x + y + z)^2\) which means that \(\displaystyle \L
x + y + z = \pm 9\).

Using those I find two ordered triples that are solutions
I cannot prove that those are all there are.

 

[matt8441]

Oh cool. I had multiplied them out and added but didn't realize it was a square. What exactly is the difference between ordered and unorder triplets?

 

[pka]

The very statement of the question implies that y is the 'middel' term.
See (x,y,z) is different from {x,y,z}.


For the third question the number \(\displaystyle \L
2^{10} \cdot 3^6 = 746496\) has 77 divisors.

Thus there are such numbers.

 

[matt8441]

The very statement of the question implies that y is the 'middel' term.
See (x,y,z) is different from {x,y,z}.


For the third question the number \(\displaystyle \L
2^{10} \cdot 3^6 = 746496\) has 77 divisors.

Thus there are such numbers.

How do you determine that that has 77 divisors? Or where did you get that number?

Edit: ohh I think I get it...

10+9+8+...+1 + 6+5+...+1 + 1 = 77?

 

[pka]

Given a number N, to find the numbers of factors write the prime fractionation.
Best to look at an example: say \(\displaystyle \L
N = 2^4 \cdot 3^6 \cdot 5^2 \cdot 7\).

Any factor of N looks like \(\displaystyle \L
2^a \cdot 3^b \cdot 5^c \cdot 7^d \quad \Rightarrow \quad 0 \le a \le 4,\,0 \le b \le 6,\,0 \le c \le 2,\,0 \le d \le 1\).
Thus a has 5 possible values, b has 7, c has 3 and d has 2.
So there are (7)(5)(3)(2) factors of N.

 

[matt8441]

Oh ok, I got that. I just need the one with the 6 sided die.

 

[soroban]

Hello, matt8441!

How many (unordered) triplets of real numbers {x,y,z} are there such that:
\(\displaystyle \;\;\begin{array}{ccc}[1]\;x(x\,+\,y\,+\,z)\:=\:26\\ [2]\;y(x\,+\,y\,+\,z)\:=\:27 \\ [3]\;z(x\,+\,y\,+\,z)\:=\:28\end{array}\)

From [1]: \(\displaystyle \L\,x\,+\,y\,+\,z\:=\:\frac{26}{x}\;\) [4]

From [2]: \(\displaystyle \L\,x\,+\,y\,+\,z\:=\:\frac{27}{y}\;\) [5]

From [3]: \(\displaystyle \L\,x\,+\,y\,+\,z\:=\:\frac{28}{z}\;\) [6]


From [4] and [5], we have: \(\displaystyle \L\,\frac{26}{x}\,=\,\frac{27}{y}\;\;\Rightarrow\;\;y\,=\,\frac{27}{26}x\;\) [7]

From [4] and [6], we have: \(\displaystyle \L\,\frac{26}{x}\,=\,\frac{28}{z}\;\;\Rightarrow\;\;z\,=\,\frac{28}{26}x\;\) [8]


Substitute into [1]: \(\displaystyle \L\,x\left(x\,+\,\frac{27}{26}x\,+\,\frac{28}{26}x\right)\:=\:26\;\;\Rightarrow\;\;\frac{81}{26}x^2\,=\,26\;\;\Rightarrow\;\;x^2\,=\,\frac{26^2}{9^2}\)

\(\displaystyle \;\;\)Hence: \(\displaystyle \L\,x\:=\:\pm\frac{26}{9}\)


Substitute into [7]: \(\displaystyle \L\,y\:=\:\frac{27}{26}\left(\pm\frac{26}{9}\right)\:=\:\pm3\)

Substitute into [8]: \(\displaystyle \L\,z\:=\:\frac{28}{26}\left(\pm\frac{26}{9}\right)\:=\:\pm\frac{28}{9}\)


Therefore, a solution is: \(\displaystyle \L\,(x,y,z)\;=\;\left(\pm\frac{26}{9},\,\pm3,\,\pm\frac{28}{9}\right)\)

 

[matt8441]

Hello, matt8441!

How many (unordered) triplets of real numbers {x,y,z} are there such that:
\(\displaystyle \;\;\begin{array}{ccc}[1]\;x(x\,+\,y\,+\,z)\:=\:26\\ [2]\;y(x\,+\,y\,+\,z)\:=\:27 \\ [3]\;z(x\,+\,y\,+\,z)\:=\:28\end{array}\)

From [1]: \(\displaystyle \L\,x\,+\,y\,+\,z\:=\:\frac{26}{x}\;\) [4]

From [2]: \(\displaystyle \L\,x\,+\,y\,+\,z\:=\:\frac{27}{y}\;\) [5]

From [3]: \(\displaystyle \L\,x\,+\,y\,+\,z\:=\:\frac{28}{z}\;\) [6]


From [4] and [5], we have: \(\displaystyle \L\,\frac{26}{x}\,=\,\frac{27}{y}\;\;\Rightarrow\;\;y\,=\,\frac{27}{26}x\;\) [7]

From [4] and [6], we have: \(\displaystyle \L\,\frac{26}{x}\,=\,\frac{28}{z}\;\;\Rightarrow\;\;z\,=\,\frac{28}{26}x\;\) [8]


Substitute into [1]: \(\displaystyle \L\,x\left(x\,+\,\frac{27}{26}x\,+\,\frac{28}{26}x\right)\:=\:26\;\;\Rightarrow\;\;\frac{81}{26}x^2\,=\,26\;\;\Rightarrow\;\;x^2\,=\,\frac{26^2}{9^2}\)

\(\displaystyle \;\;\)Hence: \(\displaystyle \L\,x\:=\:\pm\frac{26}{9}\)


Substitute into [7]: \(\displaystyle \L\,y\:=\:\frac{27}{26}\left(\pm\frac{26}{9}\right)\:=\:\pm3\)

Substitute into [8]: \(\displaystyle \L\,z\:=\:\frac{28}{26}\left(\pm\frac{26}{9}\right)\:=\:\pm\frac{28}{9}\)


Therefore, a solution is: \(\displaystyle \L\,(x,y,z)\;=\;\left(\pm\frac{26}{9},\,\pm3,\,\pm\frac{28}{9}\right)\)

Ah excellent. I had set the terms such as 26/x and 27/y equal to each other, but hadn't solved for y and z like you did. Thanks, that makes a lot of sense now.