math
posted by KMTRENd .
 Using each of the ten digits 0,1,..9 just once, is it possible to form positive integers whose sum is exactly 100?
 The only timepiece owned by a man was a wall clock which had stopped because he had forgotten to wind it. In the afternoon, he wound it while it was showing (Incorrectly) 1:00, the walked to a friend's place and noted that the correct time was 3pm. He left his friend's place at 5:30pm that evening, taking the same route home and walking at the same average speed. On arrival his clock showed 5:30. What (correct) time did he arrive home?
THANKS SO MUCH!!!!! =D

This should give you a good start.
Take the digits 19 in order and place the symbols + &  so that the sum is 100.
Example:
123+4567+89=100
The problem is often stated without any indication as to what arithmetic operations are permissible in seeking the goal, i.e., addition, subtraction, multiplication, or division, as well as exponents, parentheses, factorials, etc. It is sometimes very specific as in:
1 Arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9, in that order, using each only once, and with the aid of plus and minus signs, as desired, to produce a sum of 100.
2Arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9, in that order, using each only once, and with the aid of plus, minus, multiplication, and division signs, as desired, to produce a sum of 100.
3Arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9, in any order, using each only once, and with the aid of plus and minus signs, as desired, to produce a sum of 100.
4Arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9, in any order, using each only once, and with the aid of plus, minus, multiplication, and division signs, as desired, to produce a sum of 100.
I hope you are interested in pursuing this in some logical, or analytical, way as opposed to the time consuming trial and error process. It is so much more rewarding to wind you way through the possibilities with logic and end up with the definitive answers. Toward that end, I offer you some of my previous thoughts on the problem.
Without any specific constraints, the usual first inclination is to try it with addition only. At first glance however, it is clear that the sum of the digits 1  9 cannot add up to any more than 45. At second glance, it becomes obvious that two, or more, of the digits must be combined into one, or more, two digit numbers in order to bring the total up to 100. But can the 9 digits be made to total 100 using two digit numbers with addition only?
Lets take a somewhat analytical look at the problem.
First, note that when you add 2 even numbers, you get an even number. When you add an even number and an odd number, you get an odd number. When you add 2 odd numbers, you get an even number. Do you see what I am getting at here? You have 4 even numbers and 5 odd numbers. No matter how you add up the 5 odd numbers, either individually, or as the second digit of a two digit number, you will always end up with an odd number. Adding two pairs of
the 5 odd numbers leaves you with an odd number left over. Creating 1, 2, 3, or 4 two digit odd numbers that end with the odd digits, and add up to even numbers, you still have the one odd number left over. If you use the odd numbers as the first digit of two digit even numbers, the most you can create is four, and you still have one odd number left over. It would appear therefore, that no matter how you use the 5 odd digits, either individually or
as the first or second digit of a 2 digit number, you will always have to add the last odd number to an even sum making an odd total. Correct? Not quite.
I deliberately glossed over the one possiblity that, on the surface, appears as though it might just produce an answer for us. What if we take one odd digit, make it the first digit of a two digit number ending in an even digit. We then have four odd digits remaining, the sum of which will be even, plus the two digit even number, plus the three other even digits. Have we stumbled onto the secret here? Lets see.
What we are really trying to do is take our given sum, i.e.,
9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 or
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 whatever you choose,
subtract one odd digit, X, and one even digit, Y, combine these digits into a two digit number of the form 10X + Y, add it to the sum of the remaining digits, and have it equal 100. This translates into
45  X  Y + [10X + Y] = 100
Expanding and simplifying, we end up with
45  X  Y + 10X + Y = 100 or 45  X + 10X = 100 or
9X = 55
making the choice of the even digit irrelevant, and the supposed required odd digit being X = 6.1111.....!
I guess this makes it clear that it is impossible, as the number is neither odd, nor an integer. Making a few trial runs on either side of our answer will soon convince you that it is impossible.
[45  5  6 + 56 = 90] [45  7  6 + 76 = 108]
[45  5  4 + 54 = 90] [45  7  4 + 74 = 108]
[45  5  2 + 52 = 90] [45  7  2 + 72 = 108]
but, just to verify the validity of the oddball answer
[45  6.11111  6 + 67.11111 = 100
[45  6.11111  4 + 65.11111 = 100
[45  6.11111  2 + 63.11111 = 100
Thus, the individual digits, 1 through 9, or any combination of one or two digit numbers using 1 through 9, cannot be made to add up to 100 using addition only.
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What if we now consider the use of subtraction, as well as addition. It quickly becomes clear that it can be done, though, one or more, two digit numbers are required and we obviously must create a two digit number that, when added to the remaining single digit numbers, adds up to more than 100 in the first place from which we subtract some number, to equal 100. Can we zero in on any specific solutions with the approach taken above? Lets take a
look.
First, it is worth noting that we don't necessarily have to be concerned with the odd number of odd digits anymore as we can now obtain an initial odd sum, greater than 100, from which we can subtract another odd number in order to arrive at our target of 100. Using the above approach, and letting Z equal the digit we subtract at the end, we seek the solution of
45  X  Y  Z + [10X + Y]  Z = 100 from which we get
45  X  Y  Z + 10X + Y  Z = 100 from which we get
9X  2Z = 55 or
2Z = 9X  55
By inspection, X must be equal to, or greater than, 7, thus for X = 7 & Z = 4, (X = 8 & Z = 8.5, and X = 9 & Z = 13), making X = 7 our only viable solution. Therefore, since the Y's cancel out in our above expression, we can subtract any digit for Y and get the same result:
X = 7 Y = 1 Z = 4 9 + 8 + 6 + 5 + 3 + 2 + 71  4 = 33 + 71  4 = 100
Y = 2 9 + 8 + 6 + 5 + 3 + 1 + 72  4 = 32 + 72  4 = 100
Y = 3 9 + 8 + 6 + 5 + 2 + 1 + 73  4 = 31 + 73  4 = 100
Y = 5 9 + 8 + 6 + 3 + 2 + 1 + 75  4 = 29 + 75  4 = 100
Y = 6 9 + 8 + 5 + 3 + 2 + 1 + 76  4 = 28 + 76  4 = 100
Y = 8 9 + 6 + 5 + 3 + 2 + 1 + 78  4 = 26 + 78  4 = 100
Y = 9 8 + 6 + 5 + 3 + 2 + 1 + 79  4 = 25 + 79  4 = 100
So much for the use of a single two digit number.
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Is it possible to accomplish our goal with two, two digit numbers? Lets see.
We first have
45  W  X  Y  Z + [10W + X]  [10Y + Z] = 100 or
45  W  X  Y  Z + 10W + X  10Y  Z = 100 or
45 + 9W  11Y  2Z = 100 or
9W  11Y  2Z = 55
Dividing through by the lowest coefficient, we get
4W + W/2  5Y  Y/2  Z = 27 + 1/2 which reduces to
(W  Y  1)/2 = 27  4W + 5Y + Z
Since we are seeking integer answers, (W  Y  1)/2 must be an integer, k, resulting in
W  Y = 2k + 1 and W = 2k + Y + 1 and Y = W  2k  1
Substituting back into 9W  11Y  2Z = 55 yields 18k + 9Y + 9  11Y  2Z = 55 which reduces to
Y + Z = 9k  23 and Y = 9k  Z  23
Equating the Y's we get
W  2k  1 = 9k  Z  23 or W + Z = 11k  22
By inspection of W + Z = 11k  22, k must be equal to, or greater than, 3.
By inspection of Y + Z = 9k  23, k must be less than 4. Therefore, k = 3 is our only solution. That was easy.
For k = 3, we then have the following possibilities.
k = 3 Possibilities
W + Z = 11 2+9, 3+8, 4+7, 5+6, 6+5, 7+4, 8+3, 9+2
W  Y = 7 92, 81
Y + Z = 4 1+3, 3+1
If Y + Z = 1 + 3, W  Y can only be 8  1 making Y = 1, Z = 3, and W = 8.
If Y + Z = 3 + 1, niether W  Y candidate is possible, so Y + Z cannot be 3 + 1 and Y + Z = 1 + 3 is the only solution for k = 3. It also tells us, as before, that X can be any unused digit, giving us six solutions for the k = 3 case, namely;
X = 2 [9 + 7 + 6 + 5 + 4 + 82  13 = 100]
X = 4 [9 + 7 + 6 + 5 + 2 + 84  13 = 100]
X = 5 [9 + 7 + 6 + 4 + 2 + 85  13 = 100]
X = 6 [9 + 7 + 5 + 4 + 2 + 86  13 = 100]
X = 7 [9 + 6 + 5 + 4 + 2 + 87  13 = 100]
X = 9 [7 + 6 + 5 + 4 + 2 + 89  13 = 100]
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Might there be solutions involving three two digit numbers? I'll leave that for you to explore.
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Is it possible to derive solutions with a three digit number and a two digit number, plus the remaining single digit numbers, totaling 100, using addition and/or subtraction? Lets go and see where it takes us.
If we create a three digit number, we must then, as a minimum, subtract one two digit number to get back down to 100, if our three digit number falls between 110 and 199. This means we would have to subtract five digits from our basic sum of 45 for the nine digits, leaving only four digits to initially add. Taking the same approach, we derive our initial general expression of
45  X  Y  Z  V  W + [100X + 10Y + Z]  [10V + W] = 100
Simplifying  99X + 9Y  11V  2W = 55
For three digit numbers falling between 110 and 199, X = 1 yielding
11V + 2W  9Y = 44
Dividing through by 2 yields
5V + V/2 + W  4Y  Y/2 = 22 or (Y  V)/2 = W + 5V  4Y  22
Again, (Y  V)/2 must be an integer k from which
Y  V = 2k or Y = 2k + V and V = Y  2k.
Substituting V = Y  2k into 11V + 2W  9Y = 44 and simplifying, we get
2Y + 2W = 44 + 22k from which
Y + W = 22 + k and Y = 22 + k  W
Equating the Y's we get
2k + V = 22 + k  W or V + W = 22  k
Since the maximum (Y + W) can be is 17, no k in Y + W = 22 + k is valid.
Therefore, if I didn't make a slip someplace, there is no three digit solution between 110 and 199.
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What if we consider two three digit numbers. We're off and running.
We start out with
45  U  V  W  X  Y  Z + (100U + 10V + W)  (100X  10Y  Z) = 100
which leads us to 99U + 9V  101X  11Y  11Z = 55
Dividing through by 9 and simplifying yields
(2X + 2Y + 2Z + 1)/9 = 11U + V  9X  Y  Z  6
Again, (2X + 2Y + 2Z +1)/9 = k from which we get X + Y + Z = (9k  1)/2
By inspection, k must again be odd and no more than 5. With k = 5, the maximum X + Y + Z = 22.
For k = 5, X+Y+Z = 22 = 9+8+5, 9+7+6, 9+6+7, 9+5+8, 8+9+5, 8+6+9
k = 3, = 13 = 7+5+1, 7+4+2, 7+2+4, 7+1+5, and so on.
Well, this could take some time but I think that you get the idea by now. The only reason I took you through this exercise was for the purpose of showing you how it is possible to analytically derive the singularly unique solutions to the problem. I leave it for you to pursue this road on your own, if you desire, to see how many viable solutions you can derive.
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Another avenue of approach is to make use of addition and multiplication alone. Lets see where a trial run takes us.
We start out with a basic expression of
45  X  Y + XY = 100 which leads to
X = (55 + Y)/(Y  1) = an integer
A few quick trials leads us to the fact that Y can only be 8 or 9, resulting in X = 9 and/or 8 respectively, representing one solution. So we have
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8x9 = 100 How elegant.
Why don't you explore solutions having two products, and possible using subtraction also. I think you will be pleasantly surprised as to how easy it becomes to zero in on the viable solutions. Let me know how you make out.
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As for the constraint that the numbers remain in their original order, there are a large number of different ways in which the arithmetical signs may be placed between the nine digits, arranged in numerical order, so as to give an expression equal to 100, some of which were derived earlier. Very often, the solutions involve not only the arithmetic signs and symbols but parentheses, factorial signs, etc. After a while, the problem began to be
posed in such a way as to require the least number of signs, symbols, strokes, parentheses, etc., the goal being to derive the singularly simplest solution having the least number of signs and symbols. Some that have surfaced over the years are given below:
1 1 + 2 + 3 + 4 + 5 + 6 + 7 + (8X9) = 100 Looks familiar !
2 (1X2)  3  4  5 + (6x7) + (8X9) = 100
3 1 + (2X3) + (4x5)  6 + 7 + (8x9) = 100
4 (1 + 2  3  4)x(5  6  7  8  9) = 100
5 (1 +(2x3) + 4 + 5 + 67 + 8 + 9 = 100
6 (1X2) + 34 + 56 + 7  8 + 9 = 100
7 12+ 3  4 + 5 + 67 + 8 + 9 = 100
8 123 4  5  6  7 + 8  9 = 100
9 123 + 4  3 + 67  8  9 = 100
10 123 + 45  67 + 8  9 = 100
11 123  45  67 + 89 = 100 WOW
You might also be surprised to see that it can be done in the reverse also.
1 98  76 + 54 + 3 + 21 = 100
2 9  8 + 76 + 54  32 + 1 = 100
3 9  8 + 7 + 65  4 + 32  1 = 100
4 9  8 + 76  5 + 4 + 3 + 21 = 100
5 98  7  6  5  4 + 3 + 21 = 100
6 9 + 8 + 76 + 5 + 4  3 + 2  1 = 100
7 9 + 8 + 76 + 5  4 + 3 + 2 + 1 = 100
8 98 + 7 + 6  5  4  3 + 2  1 = 100
9 98 + 7  6 + 5  4  3 + 2 + 1 = 100
10 98 + 7  6 + 5  4 + 3  2  1 = 100 To name a few.
I trust you have found this interesting, informative, and at the very least, amusing. 
thanks nice answer :)

that wasn't the right answer. the question asks for "the sum", meaning the minus sign can't be used
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