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

### I am 5 digits long...

```
From: Anonymous
Date: Fri, 16 Dec 94 10:43:05 EST
Subject: Math Problem

Hello This is Mario Ceste and Jeff Larioni from Floris
Elem. School in Herndon, Virginia. Our Problem is this

I am 5 digits long

I am divisible by 3 and 9, but not 6

My digits add up to 27

When my first and last digits are added, you will
get a multiple of seven.

My lowest digit is 3

If you add up my second and fourth digits, you get 10

No number is used more than once

On the left of the comma, the numbers from left to
right decrease by 1.  On the right of the comma, the numbers
increase by 3.
```

```
Date: Fri, 16 Dec 1994 11:40:41 -0500 (EST)
From: Dr. Sydney
Subject: Re: Math Problem

Hello Mario and Jeff!

Thanks for writing Dr. Math.  This is quite a big word problem --
pretty tricky.  Let's see...  I'll tell you how I approached it, though
there are many different ways of doing a problem like this one.

Since the lowest digit is three and the numbers to the right of the comma
increase by three and the numbers to the left of the comma decrease by
one, the three must either be the digit in the thousands place or the digit
in the hundreds  place, right?

So, we can check both possibilities:

If the three is in the thousands place, there must be a 7 in the tens place,
since the digits in the thousands place and the tens place must add up to
10.  But that is impossible, since the digit in the ones place is supposed
to be 3 more than the digit in the tens place -- if 7 is in the tens place,
then that would mean 10 is in the ones place, but we can't have a two digit
number in the ones   place.  So, the  three must not in the thousands place.

Consider the other possibility (and we'll hope it works!).  If the three is
in the hundreds place, then we have a 6 in the tens place and a 9 in the
ones place.  Since there is a 6 in the tens place, and the digits in the
tens place and the thousands place have to add up to 10, there must be a
4 in the thousands place.  This means that the ten thousands place must
have a 5 (one bigger than 4).  So, our number is this:

54,369

Now, check all of the properties.  Do the digits add up to 27?  Is this
number divisible by 9 and 3 but not 6? etc.

I think you'll find this number works!

So, there you have it.  If you have any other questions, please write back.

--Sydney

P.S.  We are so busy answering students' questions that we usually don't
take questions like this that are more of a puzzle than anything else, so if
you write back, try not to ask any "puzzle" questions.  If puzzles are what
interest you (which is Great!--they can be lots of fun) you might want to
subscribe to the newsgroup rec.puzzles on the internet.  Have fun with it!
```
Associated Topics:
Elementary 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
Math Forum Home || Math Library || Quick Reference || Math Forum Search