SAT problem

• May 15th 2012, 03:57 PM
klg
SAT problem
the question is:
A four-digit integer, WXYZ, in which W,X,Y and Z each represent a different digit, is formed according to the following rules.
1. x=w+y+z
2. w=Y+1
3. Z=W-5
what is the four-digit integer?

I have tried every system of equation and substitution I can think of!
• May 15th 2012, 04:15 PM
emakarov
Re: SAT problem
Obviously, you have four variables and three equations, so you will not be able to find a unique solution using regular methods for solving systems of equations. You have to use the facts that digits are integers between 0 and 9.

We have x = 3(y - 1) <= 9. On the other hand, z = y - 4 >= 0, so y >= 4. This gives a unique value for y, from which other digits can be found.

In mathematics (as well as in physics and programming), it's not good to denote the same variable by a lower- and uppercase letters.
• May 15th 2012, 04:23 PM
skeeter
Re: SAT problem
Quote:

Originally Posted by klg
the question is:
A four-digit integer, WXYZ, in which W,X,Y and Z each represent a different digit, is formed according to the following rules.
1. x=w+y+z
2. w=Y+1
3. Z=W-5
what is the four-digit integer?

I have tried every system of equation and substitution I can think of!

w = y+1
z = w-5 = (y+1) - 5 = y-4 ... this says y > 4

x = w+y+z = 3y-3 using the above equations

x = 3y-3 implies that y has to be one of {1,2,3,4} ... look two lines up for the additional restriction.

you should be able to figure it out from here ...
• May 15th 2012, 04:36 PM
klg
Re: SAT problem
Got it.
Thanks!
• May 17th 2012, 02:59 PM
Soroban
Re: SAT problem
Hello, klg!

This problem doesn't require fancy algebra . . .

Quote:

A four-digit integer, WXYZ, in which W,X,Y and Z each represent a different digit,
is formed according to the following rules.

. . $[1]\;X\,=\,W+Y+Z$
. . $[2]\;W\,=\,Y+1$
. . $[3]\;Z\,=\,W-5$

What is the four-digit integer?

From [2]: . $Y \,=\,W-1$
From [3]: . $Z \,=\,W-5$

Substitute into [1]: . $X \:=\:W + (W-1) + (W-5) \quad\Rightarrow\quad X \:=\:3W - 6$

We have: . $\begin{Bmatrix}(a) & W &=& W \\ (b) & X &=& 3W - 6 \\ (c) & Y &=& W-1 \\ (d) &Z &=& W - 5 \end{Bmatrix}$

Since $X \le 9$, we have: . $(b)\;3W-6 \,\le\,9 \quad\Rightarrow\quad W \,\le\,5$

Since $Z \ge 0$, we have: . $(d)\;W-5 \,\ge\,0 \quad\Rightarrow\quad W \,\ge\,5$

Therefore: . $\begin{Bmatrix}W &=& 5 \\ X &=& 9 \\ Y &=& 4 \\ Z &=& 0 \end{Bmatrix}$
• May 17th 2012, 03:43 PM
klg
Re: SAT problem
That is WONDERFULLY simple, and seems so clear now!
Thank you so much!
klg