Divisibility by 37
Take a number of 3 figures, and add to that its "rotation":
257 + 572 + 725 = 1554
The "rotation sum" can be divided by 37. (1554 = 42 x 37)
Take a look at this one:
899 + 998 + 989 = 78 x 37
360 + 603 + 036 = 27 x 37
Prove that this can be said for every 3-digit number.
The number 37 can be expressed as 111/3, so
16 x 37 =
16 x 111/3 =
(5 + 1/3) x 111
= 592
= 555 + 37
In fact all 3-digit multiples of 37 will be of the two possible forms:
aaa + 37
or
aaa - 37
Now if you take a number like
order of the digits, you get
592 = 500 + 90 + 2
259 = 200 + 50 + 9
and rotate the
Now to get this from the original you +300 + 40 - 7 = 333, so we have:
259 = 592 - 333
= M(37) - M(37)
= M(37)
where M(37) = multiple of 37.
You can see that any rotation of the digits of a multiple of 37 is
done by either adding or subtracting 333, and since 333 is a multiple
of 37, the result remains a multiple of 37.
Check with some other multiple of 37, say
518 - 333 = 185 and 518 + 333 = 851
digits.
Another example
17 x 37 = 629
14 x 37 = 518
(= 555-37)
both being rotations of the
(= 666-37)
629 - 333 = 296
and 629 + 333 = 962
and once again these are
cyclic rotations of the original digits. The question is why this
happens.
We showed that every 3-digit multiple of 37 is of the form:
aaa + or - 37
Taking the +37 case we therefore have the digits in the following
pattern
a
a+3
a+3
a+6
a+3
a+7
a+7
a+10
now add 333
but this means carry a 1 to the tens
column, giving
a
So the digits are now
instead of
a+3
a
a+7
a+3
a
a+7
As you can see, there has been a rotation of the digits.
What happens if we have
digits are
aaa - 37 ?
a
a
a-3
a-4
So instead of digits
we get
a
a-4
a-3
a-4
a-7
a+3
and if there is a carry
now subtract 333
a-7
a+3
a
a
and applying the carry
a-4
a+3
a+3
a
and again it is a cyclic rotation of digits but now in the reverse
direction.
The fact that multiples of 37 are
for this odd pattern.
aaa + 37
or
aaa-37
is the reason
-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
ANOTHER ONE!
I don't think your last explanation was easy for many people. Therefore I
offer this solution:
Take the number 254.
Look at the rotations of its digits and note that 254 + 542 + 425 is
divisible by 37 (1221 / 37 = 33).
Now look at the general form of the number abc. (The other rotations are
bca and cba.) It is easy to see that the number abc can be written:
100a + 10b + c
and so the other rotations are
100b + 10c + a
100c + 10a + b
Adding these up gives:
100(a+b+c) + 10(a+b+c) + (a+b+c)
This form can be written as
111(a+b+c)
Note that 111 is divisible by 37.
QED.
Date: 07/30/2001 at 10:22:50
From: Doctor Peterson
Subject: Re: Divisibility by 37; another solution
My impression, looking at Dr. Anthony's original answer, is that he was
really answering a somewhat different (and harder) question: prove that
every 3-digit multiple of 37, when rotated, remains a multiple of 37.
I think that's why it's hard to follow.
Your approach to proving that the sum of all rotations of any three-digit
number is divisible by 37, is just what I would probably have done, and is
a very neat, straightforward proof.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/