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

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