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A
Bracelet of Numbers
by
Don Allen, National Aboriginal Career Symposium, Ottawa,
Canada
The
numbers that we deal with in this simple investigation
are the ten digits, 0 to 9. Choose one such number. Choose
a second, which need not be different. Obtain a new number
by adding the first two numbers, and recording only the
ones digit of the sum. Thus, for 8, 9, we have 8, 9, 7,
..., the start of an unending sequence. Obtain a "next"
number by similarly combining the final two. The sequence
becomes 8, 9, 7, 6, 3, 9, 2, 1...Continue. Predict what
may happen, and investigate
to see if you are right.
Try,
at leisure, a triple of starting digits instead of a pair.
How, otherwise, might the procedure be modified to produce
different, but interesting, results.
Hailstone
Numbers
by
Don Allen, National Aboriginal Career Symposium, Ottawa,
Canada
Known
now for over 60 years, an attractive "number"
investigation involves the so-called hailstone numbers -
you'll also see references to the Collatz conjecture and
to the "3x + 1" problem. Choose any positive
integer. The number that you choose necessarily is even
or is odd. If it is even, divide it by 2. If it is odd multiply
it by 3, then add 1. You obtain in this way a new number.
Repeat the process with the number that you obtain. Continue
to a sequence of positive whole numbers. Stop only if you
reach 1. Numbers that take you to 1 are hailstone numbers.
An
example: You choose the starting number, 15. Your sequence
is: 15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5,
16, 8, 4, 2, 1. Your sequence rises to a height of 160,
and does reach 1 - in 17 steps; 15, therefore, is a hailstone
number. Try 100 (25 steps). Try 31 (106 steps). Try 27.
Which
numbers are hailstone numbers? Are all numbers hailstone
numbers? What might happen if you were to change the rules?
Kaprekar's
Constant
by Don Allen, National
Aboriginal Career Symposium, Ottawa, Canada
This
number activity has more general possibilities, but to introduce
it we focus on its four-digit case. First, write down a
four-figure number (not all digits the same), say 3529.
Now, rearrange the digits to form the largest number that
you can (9532, I suggest) and the smallest such number (2359).
Subtract these two numbers (large minus small), and get
a new number, 7173. Repeat the process: 7731- 1377 = 6354.
Repeat again: 6543 - 3456 = 3087. And again: 8730- 0378
= 8352. And again: 8532 - 2358 = 6174. Now, observe: 7641
-1467 = 6174… again, and again! For the amateur mathematician
who discovered it, D. R. Kaprekar (he lived in India), this
number, 6174, is now known as Kaprekar's constant. Four-figure
numbers "lead to" 6174. Why? Show that 3256 leads
to 6174 in two steps, 9017 leads to 6174 in seven steps.
Would there be a Kaprekar constant for three-figure numbers?
...for five-figure numbers?
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