Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Sums of Consecutive Integers with Digital Sums

Date: 01/27/2004 at 00:20:54
From: Patrick
Subject: 100

Find all sets of positive consecutive integers that meet the following
two conditions:

1. The integers sum to one hundred.
2. The digits of the integers sum to a number greater than 30.



Date: 01/27/2004 at 16:25:16
From: Doctor Ian
Subject: Re: 100

Hi Patrick,

At first I was going to ask if this just referred to pairs of
consecutive integers, but that can't be the case, because there are
_no_ pairs like that.  

(The closest ones would be 49 + 50 = 99 and 50 + 51 = 101.)

So now we have to consider any number of integers.  Which makes this a
pretty interesting problem!

The hard thing is that there just seem to be so many possibilities. 
But is this really the case?  We've already ruled out all the pairs,
and all the integers greater than 48!  Maybe we can rule out some more
in the same way.

For example, suppose we have three consecutive numbers.  Then they're
going to have to be around 33:  

  32 + 33 + 34 =  99
  33 + 34 + 35 = 102

So there can't be any groups of three, and we don't have to worry
about any numbers greater than 31.  So we're making progress.  

Let's try groups of four.  The numbers will have to be around 25:

  23 + 24 + 25 + 26 =  98
  24 + 25 + 26 + 27 = 102

You can keep going this way, but how long will that continue?  How
large a group would you have to consider?  

Note that to get close for a group of a particular size, we can figure
out an average size by dividing 100 by our number:

  2 times an average of 50 =       100
  3 times an average of 33 = about 100
  4 times an average of 25 =       100
  5 times an average of 20 =       100
  6 times an average of 18 = about 100

and so on.  Then we're going to pick values around the average, with
about half above, and the other half below. 

What if we try something like 

  20 times an average of 5 =       100

That looks good, but if we think about it, we see that it can't work,
because we'd have to start considering negative numbers.  (Do you see
why?  About half of those values are going to have to be less than 5.) 

So there is a limit to the size of the groups we have to consider, and
it's less than 20.  

There's still some work to do, but can you take it from here?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Number Theory
High School Puzzles
Middle School Puzzles

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

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/