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Consecutive Integers, Divisible by the Sums of Their Digits

Date: 05/20/2002 at 05:20:20
From: Mike
Subject: Problem Solving

Find the first three consecutive integers greater than 10, each 
of which is divisible by the sum of its digits.

I know that the numbers are 110, 111, and 112.  But I don't 
know how to explain why they are the first numbers that work. 

Could you please help me out?
Thank you.


Date: 05/20/2002 at 13:07:59
From: Doctor Mike
Subject: Re: Problem Solving

Hi,   

It's always nice to hear from another Mike. 
   
Another word for "explain" that mathematicians use for
what you are talking about is "prove".  A proof is a logical 
sequence of steps that will demonstrate, beyond any doubt,
that the given answer is true.  This idea of proof is the
heart of mathematics.  It's good you are getting started
on it.   
  
One method of proof, which may be the best choice here, is
called the "brute force" technique. No fancy logic needed;
only straightforward observation. You can just check every
number from 11 to 112 and write down in a list which ones
are actually divisible by the sum of their digits.  The
list will start out 12, 18, 20, 21, 24, 27, 30 and so on. 
When you get to the point of adding 112 to the list, look
back on the list and observe that there were not any 3
in a row until that point.  Done!  Of course there is a
possibility you made some kind of mistake, so you should
check your work over for such things. (Even experienced
mathematicians make mistakes, but they are smart enough
to know that about themselves, and always re-check the
work they have done to make sure they have not missed
anything important.)
  
After you have done this, and in fact ANY time you prove
something, it is good to try to find a better way.  You
may find out something that allows you to explain why
it is true, but in a way that is more quick or clever.
I did this but did not come up with one here.  However,
I did notice some interesting patterns in the list.  I
saw that ANY 2-digit number which ends in the digit 
zero is automatically on the list.  Why?  Because such
a number "x0" has x+0=x as the sum of its digits, and
the number equals x times 10 so the number is divisible
by x.  
  
What other patterns are there?  When I made my list I
did it by writing the numbers down in rows and columns,
with the teens first, then the 20s in a row, then
the 30s, etc.  I circled the ones for the list, and
marked out with an X all the rest.  Geometrically, I
noticed that there was sort of a "diagonal" which had
circles around them.  On closer inspection I saw that
these were the 2-digit numbers whose digits summed to
nine.  That is 18, 27, 36, 45, 54, 63, 72, 81 and 90.
"Gee, why is that?" I asked myself.  
  
Here is an opportunity for using some real logic (and
algebra) to prove something.  Any such number is of
the form "xy" where x+y = 9.  Is xy divisible by 9?
Yes, because  
  
 xy = 10*x + y 

    = 10*x + (9-x) 

    = 10*x - x + 9 

    = 9x + 9
  
and 9x+9 factors into 9*(x+1).  Pretty slick, right?
You should go thru and tell why each "=" step is
correct.  It's not hard or tricky at all.
  
That is about the extent of the regularity that I saw,
so the most efficient and clear way to explain it seems
to be the way I tried first.  
  
I hope this has helped.
   
- Doctor Mike, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/22/2002 at 04:53:32
From: Mike
Subject: Problem Solving

Hi Doc Mike,

Thank you very much for your help with the math problem, it was very 
much appreciated. Your information really helped me out.

I have some other problems which are a bit trickier. Maybe I will 
send them to you some time.

Thanks again,
Mike
Associated Topics:
High School Basic Algebra
High School Number Theory

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