Question 205387
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'.