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five odd positive digits that equals 20.
what are they?
tricky huh ;)?

2006-09-12 05:04:58 · 14 answers · asked by tinyt00fy 1 in Science & Mathematics Mathematics

no guys i need only five digits thta's lot more than five
just think

2006-09-12 05:10:18 · update #1

carbon-based is right it is impossible if thought about it but there's a trick in the words

2006-09-12 05:14:07 · update #2

14 answers

Not tricky at all, just impossible.

The sum of an odd number of odd numbers is always odd.

So the sum of 5 odd numbers can NOT be 20.

2006-09-12 05:09:42 · answer #1 · answered by Carbon-based 5 · 0 0

trick in the words? There is missing information. Five digits that equals 20? 5*5/5+15? They are five positive digits and equals 20...

2006-09-12 12:21:53 · answer #2 · answered by dahfna 3 · 0 0

If the five digits are all positive (greater than zero), and integers, it isn't possible to meet these restrictions.

All odd integers fit the form n = 2x+1, with x = . . . -2, -1, 0, 1, 2, . . .

Mathematically, then, you need five odd integers, represented as 2v+1, 2w+1, 2x+1, 2y+1, and 2z+1 that, when totaled together must equal 20.

So you would need to have:
20 = (2v+1)+(2w+1)+(2x+1)+(2y+1)+(2z+1)
20 = 2v+2w+2x+2y+2z+5
15 = 2(v+w+x+y+z)
7.5 = v+w+x+y+z
and there's no way you could have five integers that add up to a fraction.

**However**, you can pull some tricks, like 1+3+5+11 = 20. That satisfies the "five odd digits" requirement, and reaches the required total, by playing semantic games with the word "digits".

2006-09-12 12:18:05 · answer #3 · answered by Dave_Stark 7 · 1 0

You say five odd positive DIGITS, not numbers and so I assume you mean that 11 is made up of two digits, both of which are odd.

If you meant numbers, then since Odd+Odd=Even, Even+Even=Even and Odd+Even=Odd, then

Odd+Odd+Odd+Odd+Odd

=(Odd+Odd)+(Odd+Odd)+Odd

=(Even+Even)+Odd

=Even+Odd

=Odd

20 is even, so it would be impossible.

However, continuing with the logic that by the word 'digit' you mean either 0,1,2,3,4,5,6,7,8 or 9, then you can get:

11+7+1+1=20
11+5+3+1=20
13+5+1+1=20
13+3+3+1=20
17+1+1+1=20

All these equal 20 and so are possible answers.

2006-09-12 14:23:17 · answer #4 · answered by me 2 · 1 0

I am assuming that you cannot use a number twice? I don't see how this is possible as using the lowest consecutive 5 odd whole numbers gives only 16 and using 1,3,5,7 gives 20 but thats only 4 numbers. I am not saying it can't be done but I'm certainly not seeing an obvious solution if I am understanding the rules correctly

11,1,3,5 or 1,3,7,9=20 but thats only 4 numbers


if you use larger numbers you cant use 5 before you exceed 20 so please tell me you intend to give a solution somehow so I won't lose my mind-----> hibbsj1919@yahoo.com

2006-09-12 12:17:50 · answer #5 · answered by j h 2 · 0 0

Where does the asker say 'adds up'

He says 'equals'. But clearly no odd digit 'equals' 20.

In fact no digit = 20 unless you have a base 21 or higher system.

Perhaps you may want to rephrase your answer.

2006-09-12 12:24:13 · answer #6 · answered by Anonymous · 0 0

five odd positive no. are 0+1+3+5+11 = 20
1, 3. 5,11 are odd numbers and 0 is neither odd or even so over here we can say 0 is odd no.

2006-09-12 12:19:28 · answer #7 · answered by CUTI... 1 · 0 0

17+3
15+5
13+7
11+9
19+1

2006-09-12 12:07:55 · answer #8 · answered by who be boo? 5 · 0 0

11+1+3+5
13+1+1+5
15+1+1+3
17+1+1+1

2006-09-12 13:19:07 · answer #9 · answered by bob h 3 · 0 0

(5+5+5+5)^1

2006-09-12 12:51:49 · answer #10 · answered by Anonymous · 0 0

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