If we have X,Y,Z,8 are digits for a number that have 4 digits and we arrange them in two ways:

8 x y z , z y x 8 if x>y>z and 8 x y z-z y x 8=7623 how many triplets (x ,y , z) are possible ?

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- Apr 16th 2012, 06:37 AMMhmh96How many triplets
If we have X,Y,Z,8 are digits for a number that have 4 digits and we arrange them in two ways:

8 x y z , z y x 8 if x>y>z and 8 x y z-z y x 8=7623 how many triplets (x ,y , z) are possible ? - Apr 16th 2012, 09:58 AMjsndacruzRe: How many triplets
I can't think of a shortcut method to solve this -- but working through the inequalities I can only find one triplet.

Since $\displaystyle x>y>z$ and $\displaystyle x\leq9$, notice that $\displaystyle z\leq7$ and $\displaystyle y\leq8$. So when looking at:

$\displaystyle 8 x y z$

$\displaystyle \underline{-z y x 8}$

$\displaystyle 7623$

(sorry, I can't get the numbers above to line up) you'll have to borrow when you begin your subtraction from the right hand column. Remember, when you borrow, you add 10 to z. So right from the start, you know that $\displaystyle z+10 - 8 = 3$ and so $\displaystyle z=1$.

Then it gets a little tricky. You're in the second to right-most column, i.e. $\displaystyle y - x = 2$. But you borrowed in your first step, so this becomes $\displaystyle y-1 -x = 2$. Since $\displaystyle y \leq x$, surely $\displaystyle y-1 \leq x$. So you'll have to borrow to carry out this subtraction. When you borrow, you essentially add ten, thus $\displaystyle y +10 -1 -x = 2$ and so $\displaystyle x=y+7$.

At this point, you can deduce that $\displaystyle x=9$ and $\displaystyle y=2$. This is the only possibilty, since if $\displaystyle x\leq8$, we have $\displaystyle y\leq1$. But this cannot be, since $\displaystyle z=1$ and we know that $\displaystyle y$ is larger than that. - Apr 16th 2012, 10:39 AMSorobanRe: How many triplets
Hello, Mhmh96!

Quote:

$\displaystyle \text{We have }x,y,z,8\text{ are digits for a 4-digit number and we arrange them in two ways: }8 x y z\text{ and }z y x 8$.

$\displaystyle \text{If }x>y>z\text{ and }8 x y z - z y x 8\:=\:7623\text{, how many triplets }(x ,y , z)\text{ are possible?}$

We have:

. . $\displaystyle \begin{array}{ccccc}& 8 & x & y & z \\ - & z & y & x & 8 \\ \hline & 7 & 6 & 2 & 3\end{array}$

We see immediately that $\displaystyle z = 1$

. . $\displaystyle \begin{array}{ccccc}& 8 & x & y & 1 \\ - & 1 & y & x & 8 \\ \hline & 7 & 6 & 2 & 3\end{array}$

In the second column, we see that: $\displaystyle x \:=\:y+7$

. .*Think about it!*

Hence, we have: .$\displaystyle (x,y,z) \;=\;(9,2,1),\;(8,1,1),\;(7,0,1)$

The last two triplets violate the stipulation: $\displaystyle x > y > z.$

Therefore, the answer is**one**triplet: .$\displaystyle (9,2,1)$

- Apr 17th 2012, 03:54 AMMhmh96Re: How many triplets
Thank you jsndacruz ,Soroban !