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A Number TrickSelect 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
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
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)]
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
and the sum of these last two numbers is
But, for all values of a considered, ( + v)= 9, so that 121( + v) = 1089. |
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