a three number digit is reversed such that the difference between the two numbers is written which of the following Cannot be that difference

Question

please tell me the trick to find .

Answer 1

Well, we need "the following" options to answer completely, but we might be able to say a few things.

 

Let's look at some examples.

101 - 101 = 0

201 - 102 = 99

251 - 152 = 99

271 - 172 = 99

301 - 103 = 198

302 - 203 = 99

401 - 104 = 297

402 - 204 = 198

403 - 304 = 99

 

100 - 1 = 99

520 - 25 = 495

 

Some patterns are emerging. First, it seems the middle digit doesn't matter. Secondly it seems only a handful of differences are possible. I should mention these could have all gone negative as well.

The possibilities are:

-891 891

-792 792
-693 693
-594 594
-495 495
-396 396
-297 297
-198 198
-99 99

0

 

We could try showing this.

A given number could be represented by its digits: x, y, and z (each between 0 and 9 inclusive). So, the values are

number1 = 100*x+10*y+z

number2 = 100*z+10*y+x

 

Let's consider the difference.

diff = number2-number1 = 100*z+10*y+x - (100*x+10*y+z)

 

Simplify.

diff = 99z - 99x = 99*(z-x)

 

And x and z are both integers between 0 and 9. So the difference between z and x (reprezented by (z-x)) is an integer between -9 and 9. So, we just need to enumerate 99*{-9,-8,...8,9}.