A Number Trick (2024)

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

100c + 10b + a

2.

Take the difference

100(a - c) + (c - a)

3.

and simplify to

99(a - c) = 99A Number Trick (1)

4.

where A Number Trick (2) = 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 99A Number Trick (3) as

99A Number Trick (4) = (10 . 9A Number Trick (5)) + (1 . 9A Number Trick (6))

5.

Since 2 A Number Trick (7) A Number Trick (8) A Number Trick (9) 9 by hypothesis, we know that 18 A Number Trick (10) 9A Number Trick (11) A Number Trick (12) 81 so that we may write

9A Number Trick (13) = 10A Number Trick (14) + v

6.

where A Number Trick (15) 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 . 9A Number Trick (16)) + (1 . 9A Number Trick (17)) = [10 . (10A Number Trick (18) + v)] + [1 . (10A Number Trick (19) + v)]

= 100A Number Trick (20) + 10(A Number Trick (21) + v) + v

7.

If the integer A Number Trick (22) + v does not exceed a single digit, then the number,

100A Number Trick (23) + 10(A Number Trick (24) + v) + v

is the number sought. Let us find all possible values of A Number Trick (25) + v for 2 A Number Trick (26) A Number Trick (27) A Number Trick (28) 9:

A Number Trick (29)
9A Number Trick (30)
A Number Trick (31) + 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, A Number Trick (32) + v does not exceed 9 (in fact, it equals 9 in all cases). We conclude that the number, 100A Number Trick (33) + 10(A Number Trick (34) + v) + v, is the one we sought, and correctly represents the difference, 99A Number Trick (35), as a number in base ten, with A Number Trick (36) occupying the hundreds place; (A Number Trick (37) + v), the tens place; and v, the units place. The number obtained by reversing the order of its digits is then

100v + 10(A Number Trick (38) + v) + A Number Trick (39)

8.

and the sum of these last two numbers is

100(A Number Trick (40) + v) + 20(A Number Trick (41) + v) + (A Number Trick (42) + v) = 121(A Number Trick (43) + v)

9.

But, for all values of a considered, (A Number Trick (44) + v)= 9, so that 121(A Number Trick (45) + v) = 1089.

A Number Trick (2024)
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