A Little Maths Observation Around The Number 1089

I came across 2 different maths puzzles from 2 completely different sources and noted that the answer to both was 1089.

Here`s the first:

Write down a 3-digit number where each of the digits are different.

For example, 652. Reverse the digits in this number and you get 256

Now subtract the smaller of these two numbers from the bigger one

652

-256

396

If you reverse the digits in the answer you get 693

Finally, add these together

396

+ 693

1089

So long as all three digits are different, you will always arrive at 1089. By way of a little exercise I`ll leave you to prove this fact.

(Hint: Your first 3-digit number can be generalised into the form 100a + 10b + c)

Meanwhile I noticed a puzzle posed as follows:

Find A, B, C and D from the equation below

ABCD

x9

DCBA

where A, B, C and D are all different integers between 1 and 9.

The method to working out the answer follows:

ABCD

x9

DCBA

Looking at the last part, (A x 9), if D can only be a single digit number between 1 and 9, then there is no other answer for D other than 9.

Why? Because multiples of 9 are 9, 18, 27 etc.... The only digit you can multiply by 9 by to get 9 or less is 1 (or 0).

If A was >1 (e.g. 2, then 9xA = 18 but we know D can only be a single digit between 1 and 9.) Moreover, A cannot be 0 as this would make D = 0 (when multiplying) and we cannot have 2 or more numbers the same. Therefore A must be 1 and consequently D must be 9 (since 9 x 1 = 9)

So, we now have

1BC9

x9

9CB1

Now that we know A and D, we can proceed to evaluate B and C.

B must be Zero. Why?

B can`t be 1 because A is 1 (and the puzzle states A, B, C and D are all different). B can`t be 2 (or more) because it would break D=9 when you multiply B by 9. So B can only be 0 here.

10C9

x9

9C01

Finally, considering all remaining possible values for C (2, 3, 4, 5, 6, 7, or 8) only 8 satisfies the above equation and gives 1089:

1089

x9

9801