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
100a + 10b + c |
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
100c + 10b + a | 2. |
Take the difference
100(a - c) + (c - a) | 3. |
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 | 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 | 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:
| | ![]() |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
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( | 8. |
and the sum of these last two numbers is
100( | 9. |
But, for all values of a considered, ( + v)= 9, so that 121(
+ v) = 1089.