A Number Trick
Select
a three digit number in which the first and the last digit differ by
at least two. Construct a second number by reversing the order of digits
in the first. Form a third number by taking the difference of the first
two. Reverse the order of digits in the third number to construct a
fourth, and add the third and fourth number. The result is the number
1089.
Example:
Select the number 732. Note: 7 - 2 = 5 > 2, making this number a
valid choice. Next, construct the number 237 by reversing the order
of digits. Take the difference:
732
- 237 = 495
From
495, construct the number 594, and add:
495
+ 594 = 1089.
Proof:
Let the originally selected number be represented by
where
a, b, c, are integers between 0 and 9 inclusive, and occupy the hundreds,
tens, and units places, respectively. We may assume, without loss of
generality, that a > c. Now, construct a new number by reversing
the order of digits
Take
the difference
and
simplify to
99(a
- c) = 99
|
4.
|
where
= a - c. We would like to represent this number, algebraically, as a
number in base ten; i.e., with a form similar to that shown in 1.
We begin by rewriting 99
as
Since
2
9 by hypothesis, we know that 18
9
81 so that we may write
9
= 10
+ v
|
6.
|
where
and v are integers between 0 and 9 inclusive, and occupy the tens and
units places, respectively. Thus, the Right Hand Side of eq.
5 may be rewritten as
(10
. 9)
+ (1 . 9)
= [10 . (10
+ v)] + [1 . (10
+ v)]
=
100
+ 10(
+ v) + v
|
7.
|
If
the integer
+ v does not exceed a single digit, then the number,
100
+ 10(
+ v) + v
is
the number sought. Let us find all possible values of
+ v for 2
9:
|
9
|
+ v |
2 |
18 |
9 |
3 |
27 |
9 |
4 |
36 |
9 |
5 |
45 |
9 |
6 |
54 |
9 |
7 |
63 |
9 |
8 |
72 |
9 |
9 |
81 |
9 |
Thus,
+ v does not exceed 9 (in fact, it equals 9 in all cases). We conclude
that the number, 100
+ 10(
+ v) + v, is the one we sought, and correctly represents the difference,
99,
as a number in base ten, with
occupying the hundreds place; (
+ v), the tens place; and v, the units place. The number obtained by
reversing the order of its digits is then
100v
+ 10(
+ v) +
|
8.
|
and
the sum of these last two numbers is
100(
+ v) + 20(
+ v) + (
+ v) = 121(
+ v)
|
9.
|
But,
for all values of a considered, (
+ v)= 9, so that 121(
+ v) = 1089.
|