document.write( "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 + ? \n" ); document.write( "

Algebra.Com's Answer #685371 by KMST(5328) $\"\"$ $\"About$

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We see that the numbers $\"0\"$ , $\"1\"$ , $\"2\"$ , and $\"5\"$ appear three times each.
\n" ); document.write( "A string of twelve whole numbers separated by $\"%22%2B%22\"$ and $\"%22-%22\"$ signs
\n" ); document.write( "can be considered a sum of twelve integer numbers, some positive, some negative, some zero.
\n" ); document.write( "We can change the order of terms in that sum (commutative property),
\n" ); document.write( "and we can groups numbers together (associative property)
\n" ); document.write( "as we wish to make it more convenient for our calculations.\r
\n" ); document.write( "\n" ); document.write( "You can safely place $\"%22-%22\"$ signs in front of the zeros,
\n" ); document.write( "because it does not matter if $\"0\"$ is \"added\" or \"subtracted\".
\n" ); document.write( "
\n" ); document.write( "To minimize the number of $\"%22%2B%22\"$ signs,
\n" ); document.write( "you would want then to be placed before the greatest whole numbers,
\n" ); document.write( "while placing $\"%22-%22\"$ signs in front of whole smaller numbers.
\n" ); document.write( "We know that $\"2%2B1%2B1%2B1=5\"$ <--> $\"0=5-%282%2B1%2B1%2B1%29=5%2B%28-2%29%2B%28-1%29%2B%28-1%29%2B%28-1%29\"$ .
\n" ); document.write( "We can place a $\"%22%2B%22\"$ sign in front of a $\"5\"$ ,
\n" ); document.write( "and place $\"%22-%22\"$ signs in front of every $\"1\"$ and a $\"2\"$ .
\n" ); document.write( "Having $\"red%281%29\"$ $\"%22%2B+5%22\"$ counteracting one $\"-2\"$ and three $\"-1\"$ terms
\n" ); document.write( "gives you a sum of $\"0\"$ for those terms.
\n" ); document.write( "After doing all that, there are only 3 asterisks left to be replaced,
\n" ); document.write( "one in front of $\"2\"$ , and two in front of $\"5\"$ .
\n" ); document.write( "The $\"2\"$ can take a $\"%22-%22\"$ sign to neutralize the $\"2\"$ at the start, which is positive but does not need a sign.
\n" ); document.write( "Then $\"red%281%29\"$ $\"5\"$ can take a $\"%22%2B%22\"$ sign,
\n" ); document.write( "and the other $\"5\"$ can get a $\"%22-%22\"$ sign with the total sum being zero.
\n" ); document.write( "It does not matter which $\"5\"$ gets the $\"%22-%22\"$ sign.
\n" ); document.write( " $\"2-0-1%2B5-2-0-1-5-2-0-1%2B5+=+0\"$ ,
\n" ); document.write( " $\"2-0-1-5-2-0-1%2B5-2-0-1%2B5+=+0\"$ , and
\n" ); document.write( " $\"2-0-1%2B5-2-0-1%2B5-2-0-1-5+=+0\"$ are all true,
\n" ); document.write( "and each one has only $\"red%281%29%2Bred%281%29=highlight%282%29\"$ $\"%22%2B%22\"$ signs. \n" ); document.write( "

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