Thread: How 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 ?

I can't think of a shortcut method to solve this -- but working through the inequalities I can only find one triplet.

Since and , notice that and . So when looking at:




(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 and so .

Then it gets a little tricky. You're in the second to right-most column, i.e. . But you borrowed in your first step, so this becomes . Since , surely . So you'll have to borrow to carry out this subtraction. When you borrow, you essentially add ten, thus and so .

At this point, you can deduce that and . This is the only possibilty, since if , we have . But this cannot be, since and we know that is larger than that.

Hello, Mhmh96!

.

We have:

. .


We see immediately that

. .


In the second column, we see that:
. . Think about it!


Hence, we have: .

The last two triplets violate the stipulation:


Therefore, the answer is one triplet: .

Thank you jsndacruz ,Soroban !