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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
instead of              a   a+3   a+7

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

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

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