Question 1069910: Every asterisk in the question 2*0*1*5*2*0*1*5*2*0*1*5 = 0 is to be replaced by either + or - so that the equation is correct. what is the smallest number of asterisks that must be replaced with + ?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! We see that the numbers  ,  ,  , and  appear three times each.
A string of twelve whole numbers separated by  and  signs
can be considered a sum of twelve integer numbers, some positive, some negative, some zero.
We can change the order of terms in that sum (commutative property),
and we can groups numbers together (associative property)
as we wish to make it more convenient for our calculations.
You can safely place signs in front of the zeros,
because it does not matter if is "added" or "subtracted".
To minimize the number of signs,
you would want then to be placed before the greatest whole numbers,
while placing signs in front of whole smaller numbers.
We know that  <-->  .
We can place a sign in front of a ,
and place signs in front of every and a .
Having   counteracting one  and three  terms
gives you a sum of for those terms.
After doing all that, there are only 3 asterisks left to be replaced,
one in front of , and two in front of .
The can take a sign to neutralize the at the start, which is positive but does not need a sign.
Then can take a sign,
and the other can get a sign with the total sum being zero.
It does not matter which gets the sign.
 ,
 , and
 are all true,
and each one has only  signs.
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