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Debate Score:190
Arguments:253
Total Votes:190
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 Can you get the numbers 1,2,3,4,5, 6, 7, 8,9 to total 100 just by adding together ? (185)

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Dermot(5738) pic



Can you get the numbers 1,2,3,4,5, 6, 7, 8,9 to total 100 just by adding together ?

I posted this question up on different sites in the past at one stage it caused several savage exchanges with a group of mathematicians who got totally passed oft that they couldn't solve it ; incidentally I've yet to have anyone get the correct solution....
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4 points

My answer to your question is, NO I cannot.

'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''

Dermot(5738) Clarified
2 points

That's honesty at its best ........................................

1 point

EDIT: SOLVED!

1+2+3+4+75+6+8=99

Note that only 9 remains.

We use the 9 as .9 recurring, though obviously 99+.99999= 99.99999

However we can utilize formulae to demonstrate that .9999 recurring actually equals 1

substituting .999 recurring for x we then do the following

x=.999 recurring

10x = 9.99999 recurring

10x - x = 9x = 9

1x = 1

Therefore 0.999 recurring is equal to one.

1+2+3+4+75+6+8=99

99+1=100

Dermot(5738) Clarified
1 point

One can not use the numbers more than once , if you wish to use 23 instead of 2 and 3 go ahead .

If you found the answer on google well done as I've never seen it repeated anywhere .

Post is over to me in messages and let's see your solution please

1 point

I haven't actually done the math yet but I know what's necessary for the solution. One needs to first use a decimal point and a vinculum to get a repeating decimal. I'll come back to this later and post my result here (assuming I can manage it!).

1 point

Solved it! I did need to use google to find that I needed to use a vinculum though, I'd never have got it otherwise!

1 point

OMG the 99 VEXES ME! Lol this is fun, and I hate math....so curse you for making me think in numbers....but also this is fun ;D

Dermot(5738) Clarified
1 point

You don't have to be a maths genius to get it Mint , the solution is I think a thing of beauty

Mint_tea(4644) Clarified
1 point

Ok are we ONLY dealing with addition or are we able to include negatives in it, such as -1 instead of actual 1? Or can we put some decimals in there? Otherwise I think I need a drink. 8D

1 point

That does not add up to 100. So how is your checkbook working out for you ?

Dermot(5738) Clarified
1 point

Works just fine you see I'm a genius you're not so those things never work out for you do they 😊

outlaw60(15375) Clarified
1 point

LMMFAO you are a genius that is not very likely Darwin but you are a genius in your own mind right LMMFAO

1 point

Are you asking if a sum 1+2+3+...n can be equal to 100 for some n?

No, it can't. This is not a difficult question.

The mentioned sum is equal to

S = n(n+1)

The number 100 factors as 5x5x2x2.

Among the 2 factors, n and n+1, one is even and the other one is odd (this is obvious)

So if 100 = n(n+1) for some n, you have

5x5x2x2 = n(n+1)

The only way to represent the product on the left as a multiplication of 1 odd one even factor, is 25x4.

Any other breakdown would contain two even factors.

And 25 obviously isn't equal to 4+1.

Dermot(5738) Clarified
1 point

Ah , so that translates into you cannot solve it .... ok ................

BigOats(1403) Clarified
1 point

I proved that it's impossible to get 100 by adding consequtive numbers starting at 1. If that's not what you were asking then be more specific. Maths is not about guessing games.

1 point

Technically the answer to the question: "Can you get the numbers 1,2,3,4,5, 6, 7, 8,9 to total 100 just by adding together ?" is NO. Because, whether or not it is actually possible, I personally cannot add these numbers together to get 100 right now, and the question begins with "Can you?" So the correct answer is NO.

Also it's an unclear question - can you use 1 and 2 as 12? Or, as WinstonC said, could you re-use a number?

1 point

Your searing honesty as usual is admirable Mack πŸ˜ŠπŸ‘Œ

I didn't mean it to be unclear as in I wanted people to creatively exercise all options as in yes grouping numbers but no one cannot re -use a number

1 point

It's impossible.

By induction we can find a formula for the values we can get.

Starting with 1 and 2 we get a total of 3. With 1 and 2 we can also have 12 or 21. We can subtract a value from the grand total then multiply by 10 for all numbers to get the combinations added together. 3 - 1 = 2 + 1 x 10 = 12. 3 - 2 = 1 + 2 x 10 = 21. Similarly if we examine any other combination of 2 numbers we get the same results. For example, with 2 and 3 the total is 5 and we get 5 - 2 = 3 + 2 x 10 = 23 and 5 - 3 = 2 + 3 x 10 = 23.

For 3 numbers we start with 1, 2, 3. The total is 6. We do 6 - 1 = 5 + 1 x 10 = 15. 12 + 3 = 13 + 2 = 15. And, 6 - 2 = 4 + 2 x 10 = 24. 21 + 3 = 23 + 1 = 24. So, the formula holds up for 3 numbers.

For 4 numbers we can have 2 combinations of numbers. So, we start with 1, 2, 3, 4. The total is 10. We have 10 - 3 = 7 + 3 x 10 = 37. 13 + 24 = 14 + 23 = 1 + 2 + 34 = 37.

By induction we see that the formula for determining the values you can get by adding up all numbers together as you have suggested. The sum of the numbers in the debate is 45. With the formula we know that the numbers can add up to 45 -6 = 39 + 6 x 10 = 99 or 45 - 7 = 38 + 7 x 10 = 108.

100 is not a possible value with the formula. Therefore, the mathematicians who are upset with you are correct.

Dermot(5738) Clarified
1 point

Correct to be upset ? Rather sensitive souls aren't they ? Thankfully a lot of them enjoy working out creative solutions to problems such as this , either way I care not about the hurt feelings of sensitive mathematicians .

Cartman(18192) Clarified
2 points

They were correct that you are an idiot.

1 point

Yeah I found out it's impossible, only just now, because I actually believed Dermot that there was a solution. And was looking for it instead of checking out if it's possible in theory...What a fucking troll!

Dermot(5738) Disputed
1 point

Another upset mathematician , so when you cannot solve something you call the questioner a troll .... interesting

BigOats(1403) Disputed
1 point

You're correct about it being impossible, but wrong about several things.

There are a lot more possible combinations and, for these combinations, a lot more possible values for the sum. For example, with 9 digits, you can group some of them into 2-digit numbers, while adding the rest as single digit nubmers.

Your proof does not factor in all these possibilites.

Whats important is that no matter how the 9 digits are grouped into 2-digit numbers, and no matter how many of them are left as single digit numbers, the resulting sum is always of the form 45 + 9n, where n is a certain integer (actually n is the sum of the "second" digits).

And this kind of number can never be equal to 100, because if it was we could conclude that 55 is divisible by 9.

The numbers can add up to 54, 63, 72, 81, 90, 99, 108 and 117

Dermot(5738) Disputed
1 point

That makes you's both wrong as it's possible .......................

1 point

I left out a bunch of calculations that were implied because of induction. I picked the 2 values that were closest to the target value of 100 to show there is a value smaller and larger that don't work. It was not meant to be a comprehensive list of the totals that could be achieved.

BigOats(1403) Clarified
1 point

It's actually possible in notation base 12 .

BigOats(1403) Clarified
1 point

It's possible just not in the decimal notation. The debate description doesn't mention that a certain notation must be used.

Over base 12:

1+2+4+5+6+7+8+93 = 100

Dermot(5738) Clarified
1 point

Tough luck Big Oaf ........................................ ,...

BigOats(1403) Clarified
1 point

But the phucking moron thinks this is "nonsense".

He thinks there's a solution in decimal notation.

1 point

That's awesome. Good job figuring that out.

1 point

The answer is: it's impossible.

The sum of all the digits 1+2+3+4+5+6+7+8+9 = 45 = S

There cannot be any 3-digit terms in the solution, only 2-digit ones at most.

When we introduce a 2-digit number, let's say with the second digit being n

We substract n from the sum of 45, and add it as 10n, so in fact adding 10n - n = 9n to the sum

If the "solution" contains 2-digit numbers as terms, the first digits of these numbers being n1, n2, ...n3 e.t.c

Then the resulting sum is:

45 + (n1 + n2 +n3+ ....)x9

And it has to be equal to 100

So 45 + (n1 + n2 +n3+...)x9 = 100

Which gives

(n1 +n2 +n3 +...)x9 = 55

And this is impossible since 55 does not divide evenly by 9.

Dermot(5738) Clarified
1 point

Yes I know you cannot solve it , I may try you on another teaser in the future

BigOats(1403) Clarified
1 point

I've just solved it and proved to you that there's no such combination of terms to give the sum 100. Furthermore, I've establised a neccesary and sufficient condition for when this is possible for the total sum S:

S minus the sum of the digits used must be divisible evenly by 9.

This is called maths. Your trolling is called bullshit.

(I really hope you're just trolling and not actually a challenged person).

1 point

51 + 3 + 4^2 + 6 + 7 + 8 + 9 = 100

This is probably the answer you came up with. It doesn't fit your scenario, but I am sure you will claim it does.

Dermot(5738) Disputed
1 point

This is you just you saying that . I don't have a ' scenario ' , but I'm claiming nothing I leave the ' claims ' to you

Cartman(18192) Disputed
1 point

Ok, so you are going to claim that my solution is wrong or not?

1 point

19+28+37+4+5+(++6)=100 as per programming it's crt it only uses + and requires all the digits just not in mathematical way but it's crt and satisfies u r question

Dermot(5738) Clarified
1 point

You must use all the numbers ............................................

Hars(2) Clarified
1 point

19+28+37+4+5+(++6)=100 as per programming it's crt it only uses + and requires all the digits just not in mathematical way but it's crt and satisfies u r question

Cartman(18192) Disputed
1 point

Which number was he missing?

1 point

Do I have to use all the numbers? Is it only 1-9 or can i create a combination like 9 and 1 for 91 and then ad 91+4+5=100?

Dermot(5738) Clarified
1 point

Yes you must use them all ,you can group them as in 12 ,29 etc etc

1 point

this is weird, can i invert the six to get a nine? Or vice-versa?

Dermot(5738) Clarified
1 point

No Mack that's not permitted , all you may do is add the numbers together

1 point

no you can't. I tried arithmetic progression .

its kind of like 91 +14 = 105

the 100 is between 91 and 105

Dermot(5738) Clarified
1 point

Well no you can't ; I can ..................................

I'm not surprised nobody has found the correct solution because your question is ambiguous as fuck.

Dermot(5738) Disputed
1 point

Remarkable , a very simply worded question with instructions to do exactly what it says and it's now ambiguous .

Quantumhead(745) Disputed
1 point

Remarkable , a very simply worded question with instructions to do exactly what it says and it's now ambiguous

Of course it's ambiguous you intellectually backward troll. Do you mean you can repeat the numbers or just use them once? Do you mean you keep following the sequence after 9 or go back to 1? Do you mean you have to use each listed number or are they optional?

why is it mostly Americans have such a hard time with a simply worded question ?

I wouldn't know you stupid twat. I'm English.

1 point

This isn't possible in decimal notation.

It is possible in notation base 12 (and in no other notations);

1+2+4+5+6+7+8+93 (base 12) = 100 (base 12)

Dermot(5738) Disputed
1 point

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BigOats(1403) Disputed
1 point

So is that your solution? Truly the work of a genious.

On par with Einsten.

By the way, is he there with you now, or has he been transferred to another ward?