Associated Topics || Dr. Math Home || Search Dr. Math

### Divisibility by 37

```
Date: 11/08/97 at 04:22:37
From: P.A. van Renselaar
Subject: Rotation sum

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.
```

```
Date: 11/08/97 at 09:21:53
From: Doctor Anthony
Subject: Re: Rotation sum

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   592 = 500 + 90 + 2   and rotate the
order of the digits, you get    259 = 200 + 50 + 9

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  14 x 37 = 518  (= 555-37)

518 - 333 = 185 and 518 + 333 = 851    both being rotations of the
digits.

Another example   17 x 37 = 629    (= 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+7     now add 333
a+3   a+6   a+10    but this means carry a 1 to the tens
column, giving
a+3   a+7    a

So the digits are now  a+3  a+7   a

As you can see, there has been a rotation of the digits.

What happens if we have  aaa - 37 ?

digits are      a   a-3    a-7  and if there is a carry
a   a-4    a+3  now subtract 333

a-3  a-7   a    and applying the carry
a-4  a+3   a

So instead of digits    a   a-4   a+3
we get                a-4   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  aaa + 37  or  aaa-37  is the reason
for this odd pattern.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 07/25/2001 at 22:58:56
From: P.A. van Renselaar
Subject: Divisibility by 37; another solution

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

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/
```
Associated Topics:
High School Basic Algebra
High School Number Theory

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search