SOLUTION: is it possible to use the digits 0-9 in an equation only using addition to add up to 100. Rules 1. you must use each number once and only once (no repeats) 2. you can group num

Question 205387: is it possible to use the digits 0-9 in an equation only using addition to add up to 100.
Rules
1. you must use each number once and only once (no repeats)
2. you can group numbers together for example... 1 and 9 -> 19 (not adding or anything just using two of the numbers to form a 2 digit number)
3. I think that covers the rules if i seem to have left anything out let me know
pattern i've noticed
the only important thing is what the the tens places add up to
ex1a. 13+24+5+6+7+8+9+0 = 72 (a multiple of nine)
ex1b. 19+20+3+4+5+6+7+8 = 72
when the tens places add up to or equals "3" then the answer is always 72
ex2a. 12+34+5+6+7+8+9+0 = 81
ex2b. 45+1+2+3+6+7+8+9+0 = 81
when the tens places add up to ro equals "4" then the answer is always 81
hmmm... interesting
the pattern continues each time the answer is always a multiple of 9
5=90
6=99
7=108
the person who gave me this problem insists very strongly that this is possible but i have found the opposite to be true
your help would be much appreciated cause the only other way that i can think of is for my dad and i to write a program which will check every possible way the numbers can be rearranged etc.
anyways,
thx a bunch
TJ
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Hi TJ,

You're absolutely correct in saying that there is no solution here. Also, you're on to something when you say that each sum is a multiple of 9. Here's why:

If we look at the sum of 1 digit numbers from 0-9, we get

0+1+2+3+4+5+6+7+8+9 = 45

Now if we group ANY two random digits, say 3 and 7, and combine them into a two digit number, we get 37. What we've done is take 3 and 7 out and added in 37 (ie go from 0+1+2+3+4+5+6+7+8+9 to 0+1+2+37+4+5+6+8+9). So subtract 3 from 45, 7 from 45, and add in 37 to get:

45-3-7+37 = 72

which is a multiple of 9. This is no coincidence. You can pick ANY pair of digits (which we'll call 'a' and 'b') and do the same. So we'll subtract out 'a' and 'b' from 45 and then add '10a+b' back in to get:

45-a-b+10a+b = 45 + 9a = 9(5+a)

Note: two digit numbers follow the form 10t+u where 't' is the tens digit and 'u' is the units digit. Ex: 10*2 + 7 = 27

If we let k=5+a, we then get 9k which is indeed a multiple of 9. We can pick yet another pair of numbers c and d and do the same to get

9k-c-d+10c+d=9k+9c = 9(k+c) = 9m ... where I let m=k+c

Do so again (with numbers e and f) to get: 9m-e-f+10e+f=9m+9e=9(m+e)=9n ... where I let n=m+e

If we keep this up, we'll eventually run out of numbers to pull out and add up. Because we keep getting a sum that is a multiple of 9, the final answer will be in the form of 9j (where 'j' is an integer).

Since 100 is NOT a multiple of 9, but 9j is, this means that no matter how hard you try, you will never reach 100 (the closest you'll get is 99). If you allow subtraction, then you can reach 100. Could it be that the person who gave you this problem meant to allow subtraction? Or did s/he want you to reach a different number?

If you are allowed subtraction, I ran a program to see all of the possible solutions and found that subtraction was used in each case (which confirms the original answer).

Note: if the program did find a case with all addition signs, then that would clearly violate the logic used above...(but it didn't, phew...). This is why it's good to verify your answer.

Here are all the possible solutions (using subtraction of course):

123+45-67+8-9 = 100
123+4-5+67-89 = 100
123-45-67+89 = 100
123-4-5-6-7+8-9 = 100
12+3+4+5-6-7+89 = 100
12+3-4+5+67+8+9 = 100
12-3-4+5-6+7+89 = 100
1+23-4+56+7+8+9 = 100
1+23-4+5+6+78-9 = 100
1+2+34-5+67-8+9 = 100
1+2+3-4+5+6+78+9 = 100
-1+2-3+4+5+6+78+9 = 100

Note: I realize now (not sure why I didn't see this earlier) that the program doesn't rearrange the digits (if the program did rearrange the digits, there would be A LOT more cases which means that it'll take much much longer). However, I've seen other postings on this exact problem and the response has been the same: 'there is no solution'.
RELATED QUESTIONS
I need help. I must have to make an equasion using digits 0-9 and only once. It also has (answered by seawind)

Using only the digits 0,1and 9 to write a decimal in billionths. Use each digit at least... (answered by CubeyThePenguin)

Using the digits 1-9 in order, insert operation signs and grouping symbols to get the... (answered by mathslover)

Can I ask your help, please, for the following problem: Using the digits from the year... (answered by MathLover1)

How to get 100 by using digits from 1 to 9. Rules- Only addition can be used.......... (answered by Edwin McCravy)

If the digits 1 through 9 were written in order, it is possible to place plus and minus... (answered by jim_thompson5910)

Use only the digits 0,1 and 9 to write a decimal in billionth. your decimal should be... (answered by CubeyThePenguin)

How do you get the answers of 34, 38, 48, and 49 using only the numbers 1, 2, 3, and 4.... (answered by Theo)

Write out the digits from 1�9 in order. Then add some plus (+) signs and times (x) signs (answered by joyofmath)