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### Digit Reversal Trick Explained

```
Date: 03/23/2001 at 15:40:50
From: I. Kouraklis
Subject: Algebra trick/puzzle

Take a 3-digit number (e.g. 875) and subtract its reverse (578). Then,
take the result (297) and add its reverse (792). The answer is always
1089, no matter what the initial numbers are. I would like to know why
this happens. I noticed that this rule has some exceptions; it doesn't
apply to numbers with the same first and third digits, since the
subtraction gives zero.

Thank you
```

```
Date: 03/26/2001 at 14:42:06
From: Doctor Rob
Subject: Re: Algebra trick/puzzle

Thanks for writing to Ask Dr. Math.

Let the original number be ABC, that is,

100*A + 10*B + C

Here A, B and C are digits, whole numbers from 0 to 9. In order to be
a three-digit number, presumably A is not zero. Then its reverse is:

100*C + 10*B + A

Subtracting the smaller from the larger, you get:

99*A - 99*C   or   99*C - 99*A

Notice that the B's drop out. Suppose A > C, so it is the former
result you get. (The same kind of argument works for A < C; A = C is
one of the exceptions below.) Then:

99*(A-C) = 100*D + 10*E + F

Notice that 0 < A-C <= 9, so A-C is a digit. Add this to its reverse:

100*F + 10*E + D

and you get

101*(D+F) + 20*E

Now F, the units digit of 99*(A-C) is 10-(A-C). Also E, the tens
digit, is always 9, and F, the hundreds digit, is (A-C)-1. This all
works because 9 >= A-C >= 1, and:

99*1 = 099
99*2 = 198
99*3 = 297
99*4 = 396
99*5 = 495
99*6 = 594
99*7 = 693
99*8 = 792
99*9 = 891

That means that the final sum is

101*(10-A+C+A-C-1) + 20*9 = 101*9 + 180 = 1089.

Notice that the A's and C's drop out.

The exception, of course, is when A = C, and then 99*(A-C) is zero.

If a 3-digit number is chosen so that the last digit is 0, when you
reverse it, you can drop the initial zero. Thus if you start with 730,
you reverse it to 037 and drop the leading 0, giving 37, and subtract
that from 730 to give 693.  Then 693 + 396 = 1089, as before.

Another condition is that when the difference is less than 100, you
have to put on a leading zero to make it a 3-digit number before
reversing and adding. For example, if you start with 392, reverse to
293, and subtract, you get 99. Make this 099 before you reverse it to
get 990, and add 99, getting 1089 as before.

For another answer to this question in our archives, see:

Proving Something is Always So
http://mathforum.org/dr.math/problems/emma4.1.98.html

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Puzzles
Middle School Algebra
Middle School Number Sense/About Numbers
Middle School Puzzles

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