The numbers game
Question: If you add up all the digits of a number and keep on adding the
resulting digits until you are left with a single digit, then if the original
digit is divisible by 3, so is the resultant digit. Thus for 129, 1 + 2 + 9 =
12, 1 + 2 = 3. Or 81, where 8 + 1 = 9. Conversely, if the original number is not
divisible by 3, then neither is the resultant digit. Thus for 131, 1 + 3 + 1 =
5.
I have written a computer program that tested this rule right up to 1 billion
and it found no exceptions. I also saw that as the original number increased
(and, of course, if it was divisible by 3) the single-digit result increased by
3 as well: 3, 6, 9, 3, 6, 9 ad nauseam.
What is going on? Has anyone proved this for all numbers? It doesn鈥檛 seem to
work for any other prime.
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Answer: Take any number wxyz鈥� where z, y,
x, w and so on are the digits going back as far as you like in
base 10. This can be written as
z + 10y + 100x + 1000w+ 鈥�
and the sum of the digits is
z + y + x + w + 鈥�
But
z + 10y + 100x + 1000w = z +
y + x + w+鈥� + 9y + 99x + 999w
+鈥�
From this calculation you can see that the number is equal to the sum of its
digits plus a number which is divisible by 9 (or 3).
Therefore, if the sum of the digits is divisible by 3, the whole number is
also divisible by 3, and if the sum of the digits is divisible by 9, again the
whole number is divisible by 9.
If you add the digits again and again, you are repeating the same process, so
the same rule still applies.
Rosa Clements
Harrogate, North Yorkshire
Answer: This is nothing more than the repeated application of the rule taught
to me at school that the sum of the digits of any number divisible by 3 is also
divisible by 3. Therefore, it provides a quick check on divisibility.
That this is universally true is easily shown. Take a number like 315. This
can obviously be written as
(100 x 3) + (10 x 1) + 5
which can then be split into
(99 + 1) x 3 + (9 + 1) x 1 + 5
Expanding this further gives
(99 x 3) + (9 x 1) + 3 + 1 + 5
Because 99 and 9 are obviously divisible by 3, the original number can only
be divided by 3 if the sum 3 + 1 + 5 is divisible. In other words, if the sum of
the digits is divisible by 3, the original number will be too. Clearly this
argument can be generalised to numbers of more than three digits.
Because divisibility by 3 is a property of both the original number and the
sum of its digits, it follows that the process can be repeated ad
nauseam in the manner noted by your reader.
W. Smith
Computational Science and Engineering Department
CCLRC Daresbury Laboratory, Warrington
Answer: As I learned back in primary school, there are a few basic tricks
that apply regarding the divisibility of round numbers.
A number is divisible by 2 if the last digit is divisible by 2; by 3 if the
sum of the digits is divisible by 3; by 4 if the number made of the last two
digits is divisible by 4 (so 125,716 is divisible by 4 because 16 is); by 5 if
the last digit is a 5 or a 0; by 6 if the number is divisible by 2 as well as by
3; by 8 if the number made of the last three digits is divisible by 8; by 9 if
the sum of the digits is divisible by 9; and by 10 if the last digit is a 0.
Although these tricks seem to differ, most of them are a result of the same
theory.
Take the numbers that are divisible by 2. When summing up the 2-times table
we can see that the last digits follow the sequence 2, 4, 6, 8, 0 which then
repeats, so only the last digit has to be looked at. An even simpler cycle
applies to multiples of 5.
The cycle for 4 is just a little bit different. Instead of repeating every
10, it repeats every 100, because 100 is the smallest power of 10 that is
divisible by 4. In the same way, the cycle for 8 repeats every 1000.
With the cycle for 3, every time that the last digit exceeds 10 it jumps down
by 1 relative to the previous time round, while the penultimate digit increases
by 1. For example, as 9 becomes 12 the last digit jumps 1 down becoming 2 not 3,
as it was in the previous cycle of 10鈥攚hile the penultimate digit goes up
1 (from 0 to 1)
Kristian Koerselman
Heerhugowaard, The Netherlands
Answer: If you tabulate the integers like this, it quickly becomes clear why
the sums add up to 3, 6 and 9 cyclically:
1 . . . 10 . . . 19
2 . . . 11 . . . 20
3 . . . 12 . . . 21鈥�.. 3 + 9n
4 . . . 13 . . . 22
5 . . . 14 . . . 23
6 . . . 15 . . . 24鈥�.. 6 + 9n
7 . . . 16 . . . 25
8 . . . 17 . . . 26
9 . . . 18 . . . 27鈥�.. 9 + 9n
where n is a non-negative integer i.e. 0, 1, 2, 3鈥�
The fact that the digits add up to 3 or a multiple thereof if the number is
divisible by 3 is well known, and is one of the 鈥渄ivisibility criteria鈥� (see
www.lifesmith.com/mathfun.html#anchor613541).
Interestingly, the same rule also applies to 9.
Steve Durbin
Barry, Glamorgan
Answer: The principle even works for numbers that are above 10. Any number in
which the sum of the digits in the odd-numbered positions equals the sum of the
digits in the even-numbered positions (or if the two sums differ by 11) will
itself divide by 11.
For example, if you take 121, the sum of numbers in odd-numbered positions is
1 + 1 = 2 while the sum of the numbers in even-numbered positions = 2. Therefore
121 divides by 11.
Elaine Turpin
Pembrokeshire
This week鈥檚 question
Lonely planet: Although all the planets of the Solar System travel around the
Sun in the same direction, Venus actually rotates on its axis in the opposite
direction to all the other planets. My son wants to know why this is and I鈥檓
ashamed to admit that I couldn鈥檛 explain it to him. Can anyone help?
Graham Tonwal
York