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### Can the Sum of 5 Odd Numbers Equal an Even Number?

```Date: 01/05/2003 at 08:18:46
From: Charlie
Subject: Odd numbers that add to make an even number

What are 5 odd numbers which, when added, equal 32 ?

It is apparently logically impossible.

This is a riddle. A friend of mine says that there is an answer that
he has promised his father he will not reveal. He says one person he
knows has solved this riddle.
```

```
Date: 01/05/2003 at 14:43:52
From: Doctor Ian
Subject: Re: Odd numbers that add to make an even number

Hi Charlie,

It's impossible, unless you use some non-standard definition for
'odd'. Note that by definition, a number n is odd if there exists some
integer k such that

n = 2*k + 1

Given that definition, the sum of 5 odd numbers looks like this:

(2a+1) + (2b+1) + (2c+1) + (2d+1) + (2e+1)

= 2(a+b+c+d+e) + 5

= 2(a+b+c+d+e+2) + 1

which, by definition, is an odd number. So if you just stay within
standard base 10 arithmetic, there is no solution.

However, by moving outside standard base 10 arithmetic, there are at
least two ways to generate an infinite number of solutions to the
riddle.

1) We can make use of of other bases, e.g.,

3 + 3 + 3 + 3 + 5 = 17 (base 10)

= 32 (base 5)

To generalize this, suppose we have 32 in some base, B.  Then in base
10, this is equal to

3B + 2

and we want that to be equal to the sum of 5 odd numbers:

3B + 2 = 2(a+b+c+d+e) + 5

Following our pattern above, let's say that

a = b = c = d               (e.g., a=1, so 2*a+1=3)

and

e = a + 1                   (i.e., the next odd number)

Then we have

3B + 2 = 2(5a+1) + 5

3B + 2 = 10a + 2 + 5

3B = 10a + 5

B = (10a + 5)/3

So here are just a few of the infinitely many possibilities:

a         B         Odd numbers
---    ---------     -------------------
1     15/3 =  5     3, 3, 3, 3, 5
4     45/3 = 15     9, 9, 9, 9, 11
7     75/3 = 25     15, 15, 15, 15, 17

In other words, for any whole number k, the numbers

a = b = c = d = 3k+1

e = 3k+3

are four odd numbers in base 10 that add up to 32 in base

B = (10a + 5)/3

= (10(3k+1) + 5)/3

2) We can make use of modular arithmetic, i.e., the kind of
'arithmetic' that we use with clocks, where

9 a.m. + 5 hours = 2 p.m.

On a 12-hour clock, 12 is the same as 0. Similarly, on a '35-hour
clock', 35 is the same as 2, so

35 + 9 + 9 + 9 + 5 = 0 + 9 + 9 + 9 + 5

= 32

For that matter, for any positive integer k,

k*35 + 9 + 9 + 9 + 5 = 32     (modulo 35)

Of course, you can generalize this to an infinite number of moduli, so
long as you choose an odd modulus greater than 32. (A 12-hour clock
has a modulus of 12.) That is,

k*m + (any four odd numbers that add up to 32) = 32 (modulo m)

I'm sure there are other ways to generate solutions as well.

Does this help?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
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