Mistakes gave rise to yet another problem!

Algebra Level 4

In mathematics, If you do the following: $x(y+z)=xy+z$ , Then it's a big mistake.

How many ordered pairs integers $x_i ,y_i,z_i$ such that $-10\leq x,y,z \leq 10$ make the equation above true?

$x_i,y_i,z_i$ are Integers

The answer is 861.

1 solution

Mehul Arora
Jun 24, 2015

$x(y+z)=xy+z$

$xy+xz=xy+z$

$xz=z$

$z(x-1)=0$ So, Either x=1, or z=0.

Case 1:

x=1, Then, We have 21 choices for y and 21 choices for z. Hence, We get 441 $21 \times 21 \times 1$ pairs of $x,y,z$ here.

Case 2:

z=0, Then, we have 21 choices for x and 21 choices for y. Hence, We get 441 pairs of $(x,y,z)$ here.

But now, we need to subtract the cases in which we have both $z=0 \ and x=1$ , which are 21 in number.

$=882-21=861$

Answer $= \boxed {861}$

Who are you go teach India you're a Genius.

Rohan Verma - 5 years, 11 months ago

Lol, I'm No Genius. I'm just an average boy willing to learn mathematics.

I'm glad you liked my solution :)

Mehul Arora - 5 years, 11 months ago

I,m not satisfied with your logic to subtract 21. I think we subtract 21 not because we have some issue in taking $|\huge{x=1}$ and $\huge{z=0}$ $\large{simaltaneously}$ but because the ordered pairs of the form $\huge{(1,y,0)}$ are repeated twice. 21 times while we consider the permutations of x=1 and 21 times again while we consider the permutations of z=0. you may also be saying the same thing but it wasn't clear thats why i asked.

Muhammad Humaiz Anjum - 5 years, 8 months ago

@Muhammad Humaiz Anjum That's exactly what I said. It's okay to have misconceptions. :)

Keep learning! :D

Mehul Arora - 5 years, 8 months ago

Mistakes gave rise to yet another problem!

The answer is 861.

1 solution

0 pending reports