## Numbers: Their Tales, Types, and Treasures.

## Chapter 9: Number Relationships

### 9.1.BEAUTIFUL NUMBER RELATIONSHIPS

Having observed some numbers that exhibit special characteristics, we now consider noteworthy relationships among other numbers. There are pairs of numbers that partner in rather unexpected relationships. There are partnerships that lead us to refer to them as amicable numbers or friendly numbers. There are numbers that partner as triples, where Pythagorean triples are a prime example. In this chapter, we will consider these and other number relationships that add some unexpected dimensions to the beauty of numbers as they relate to one another.

It is hard to imagine that there are certain number pairs that yield the same product even when both numbers are reversed. For example, 12 × 42 = 504 and, if we reverse each of the two numbers, we get 21 × 24 = 504. The same thing is true for the number pair 36 and 84, since 36 × 84 = 3,024 = 63 × 48.

At this point you may be wondering if this will happen with any pair of numbers. The answer is that it will only work with the following fourteen pairs of numbers:

12 × 42 = 21 × 24 = 504

12 × 63 = 21 × 36 = 756

12 × 84 = 21 × 48 = 1,008

13 × 62 = 31 × 26 = 806

13 × 93 = 31 × 39 = 1,209

14 × 82 = 41 × 28 = 1,148

23 × 64 = 32 × 46 = 1,472

23 × 96 = 32 × 69 = 2,208

24 × 63 = 42 × 36 = 1,512

24 × 84 = 42 × 48 = 2,016

26 × 93 = 62 × 39 = 2,418

34 × 86 = 43 × 68 = 2,924

36 × 84 = 63 × 48 = 3,024

46 × 96 = 64 × 69 = 4,416

A careful inspection of these fourteen pairs of numbers will reveal that in each case the product of the tens digits of each pair of numbers is equal to the product of the units digits. We can justify this algebraically as follows: For the numbers *z*_{1}, *z*_{2}, *z*_{3}, and *z*_{4}, we have

*z*_{1} × *z*_{2} = (10*a* + *b*) × (10*c* + *d*) = 100*ac* + 10*ad* + 10*bc* + *bd*, and *z*_{3} × *z*_{4} = (10*b* + *a*) × (10*d* + *c*) = 100*bd* + 10*bc* + 10*ad* + *ac*.

Here *a*, *b*, *c*, *d* represent any of the ten digits: 0, 1, 2,…9, where *a* ≠ 0 and *c* ≠ 0.

Let us assume that *z*_{1} × *z*_{2} = *z*_{3} × *z*_{4}. Then,

100*ac* + 10*ad* + 10*bc* + *bd* = 100*bd* + 10*bc* + 10*ad* + *ac*,

that is, 100*ac* + *bd* = 100*bd* + *ac*, and 99*ac* = 99*bd*, or *ac = bd*, which is what we wanted to prove.

There are times when the numbers speak more effectively for themselves than does any explanation. Here the numbers are related in a rather unusual way. A visual inspection of this relationship is far better than a written one. Simply enjoy!

1^{1} + 6^{1} + 8^{1} = 15 = 2^{1} + 4^{1} + 9^{1}1^{2} + 6^{2} + 8^{2} = 101 = 2^{2} + 4^{2} + 9^{2}1^{1} + 5^{1} + 8^{1} + 12^{1} = 26 = 2^{1} + 3^{1} + 10^{1} + 11^{1}1^{2} + 5^{2} + 8^{2} + 12^{2} = 234= 2^{2} + 3^{2} + 10^{2} + 11^{2}1^{3} + 5^{3} + 8^{3} + 12^{3} = 2,366 = 2^{3} + 3^{3} + 10^{3} + 11^{3}1^{1} + 5^{1} + 8^{1} + 12^{1} + 18^{1} + 19^{1} = 63 = 2^{1} + 3^{1} + 9^{1} + 13^{1} + 16^{1} + 20^{1}1^{2} + 5^{2} + 8^{2} + 12^{2} + 18^{2} + 19^{2} = 919 = 2^{2} + 3^{2} + 9^{2} + 13^{2} + 16^{2} + 20^{2}1^{3} + 5^{3} + 8^{3} + 12^{3} + 18^{3} + 19^{3} = 15,057 = 2^{3} + 3^{3} + 9^{3} + 13^{3} + 16^{3} + 20^{3}1^{4} + 5^{4} + 8^{4} + 12^{4} + 18^{4} + 19^{4} = 206,755 = 2^{4} + 3^{4} + 9^{4} + 13^{4} + 16^{4} + 20^{4}