[wolfgang]

Give me five different even numbers with a total sum of 20.

The numbers should be all(+).

You shuold not repeat any number twice.

The numbers should be added together (only) to give us a total of 20.

Zero(0) is not included.

Edited March 30, 2011 by wolfgang

[Guest]

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2,4,4,4 and 6? I didn't repeat a number twice I repeated it thrice.

[Guest]

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4+4+4+4+4=20

Meets all criteria...

[fabpig]

  On 3/30/2011 at 6:30 PM, rlgandy said:

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2,4,4,4 and 6? I didn't repeat a number twice I repeated it thrice.

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If you're going to split hairs, you DID repeat it twice. 4,4 is repeating it once!

. But it also says 5 different numbers.

[Guest]

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2, 4, 6, 8, and 10 in base 20?

Edited March 30, 2011 by thirdtimescharm

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2, 4, 6, 8 and 10 in base 15? The ten being in the one's place.

[Guest]

Well, this is a conundrum ...

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While fractions are never considered even or odd, if the definition of an even number is (loosely speaking) evenly divisible by 2, then wouldn't a fraction like 8/10ths be even? Here's my answer: 8/10, 12/10, 2, 6, 10.

Edited March 30, 2011 by JAPrufrock

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  On 3/30/2011 at 7:21 PM, rlgandy said:

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2, 4, 6, 8 and 10 in base 15? The ten being in the one's place.

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But, in a Base 15, wouldn't these numbers (all non-fractioned numbers) be (technically) odd?

Edited March 30, 2011 by JAPrufrock

[Guest]

  On 3/30/2011 at 7:27 PM, JAPrufrock said:

Well, this is a conundrum ...

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While fractions are never considered even or odd, if the definition of an even number is (loosely speaking) evenly divisible by 2, then wouldn't a fraction like 8/10ths be even? Here's my answer: 8/10, 12/10, 2, 6, 10.

The definition of an even number is any number that can be written as 2n where n is an integer.

[curr3nt]

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Base 15 would be the first base that would work. 2+4+6+8+A=20 in base 15 Since those number occur before 10(base 15) they are still even. 10(base 15) is where the "even" numbers are odd until 20(base 15)

[Guest]

  On 3/30/2011 at 8:00 PM, curr3nt said:

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Base 15 would be the first base that would work. 2+4+6+8+A=20 in base 15 Since those number occur before 10(base 15) they are still even. 10(base 15) is where the "even" numbers are odd until 20(base 15)

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You're right ... I'm confused.

[Guest]

Using different bases

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2 + 4 + 8 + 10 (base 2) + 100 (base 2)

[Guest]

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20 written in base N will be translated in :

20= 2*N^1 + 0* N^0 = 2*N.

Now , considering that all the number we are adding are even, we can take out 2 of the sum like this:

A1= 2*a

A2=2*b

A3=2*c

A4=2*d

A5=2*e.

So the sum would look like this:

A1+A2+A3+A4+A5 = 20. applying the above calculation, we get:

2( a+b+c+d+e) = 2 * N. this is equivalent to

a+b+c+d+e = N. in here, N is the base, a,b,c,d,e are greater then 0 integers.

The smallest greater then 0 (distinct) integers are 1,2,3,4,5 which add up to 15.

So the smallest base this works in is 15, with a unique solution to the question.

From here on, you can increase the base by 1 and find number that respect the relation above.

for base 16 you will 1 set of numbers that respects it:

(1,2,3,4,6) and the even number would then be (2,4,6,8,12).

for base 17 you have 2 sets of numbers that respect it:

(1,2,3,4,7) (1,2,3,5,6) with the even number just multiply this by 2 and so on.

So the answer for this question exists in any base greater or equal then 15.

[wolfgang]

  On 3/30/2011 at 6:50 PM, thirdtimescharm said:

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2, 4, 6, 8, and 10 in base 20?

but you did an additional processe(base),you should(add) only.

[wolfgang]

  On 3/30/2011 at 7:27 PM, JAPrufrock said:

Well, this is a conundrum ...

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While fractions are never considered even or odd, if the definition of an even number is (loosely speaking) evenly divisible by 2, then wouldn't a fraction like 8/10ths be even? Here's my answer: 8/10, 12/10, 2, 6, 10.

Here,you are going to devide and add,while I said ,you should (add) only.

[wolfgang]

  On 3/31/2011 at 2:02 PM, Fesoj said:

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20 written in base N will be translated in :

20= 2*N^1 + 0* N^0 = 2*N.

Now , considering that all the number we are adding are even, we can take out 2 of the sum like this:

A1= 2*a

A2=2*b

A3=2*c

A4=2*d

A5=2*e.

So the sum would look like this:

A1+A2+A3+A4+A5 = 20. applying the above calculation, we get:

2( a+b+c+d+e) = 2 * N. this is equivalent to

a+b+c+d+e = N. in here, N is the base, a,b,c,d,e are greater then 0 integers.

The smallest greater then 0 (distinct) integers are 1,2,3,4,5 which add up to 15.

So the smallest base this works in is 15, with a unique solution to the question.

From here on, you can increase the base by 1 and find number that respect the relation above.

for base 16 you will 1 set of numbers that respects it:

(1,2,3,4,6) and the even number would then be (2,4,6,8,12).

for base 17 you have 2 sets of numbers that respect it:

(1,2,3,4,7) (1,2,3,5,6) with the even number just multiply this by 2 and so on.

So the answer for this question exists in any base greater or equal then 15.

well..there is

[plainglazed]

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2 + 4 + 6 + 8 = 20 --> five unique, non-negative, non-zero, even integers whose sum is 20?

[wolfgang]

  On 4/1/2011 at 1:22 PM, plainglazed said:

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2 + 4 + 6 + 8 = 20 --> five unique, non-negative, non-zero, even integers whose sum is 20?

well...I see only four integers..2,4,6 and 8

Don`t take it as a pure mathmatics....its a trick...

[Guest]

  On 3/30/2011 at 6:04 PM, wolfgang said:

Give me five different even numbers with a total sum of 20.

The numbers should be all(+).

You shuold not repeat any number twice.

The numbers should be added together (only) to give us a total of 20.

Zero(0) is not included.

Does (+) = positive? If not, use negatives, like...

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2, 4, 8, 10, -4

[wolfgang]

  On 4/2/2011 at 9:52 PM, typodestoyer said:

Does (+) = positive? If not, use negatives, like...

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2, 4, 8, 10, -4

+...is...+ but not(positive)...and not(negative)

[wolfgang]

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+5 EVEN