It's much closer than you think!
9 8 7 6 5 4 3 2 1 = 1 0 0
Without touching the right side of the equation, find a way to make the equation above true. You are only allowed to use mathematical operators like addition and subtraction only.
Find a possible way to make the above equation true such that you used the least number of mathematical operators. Let A denote the number of addition signs used and S denote the number of subtraction sign used. Find the value of 3 A + 4 S .
Details And Assumptions:
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As an explicit example if the equation 9 + 8 − 7 − 6 − 5 + 4 + 3 − 2 − 1 = 1 0 0 is true, then you have used 4 addition signs and 5 subtraction signs. So A = 4 , S = 5 .
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You can combine the digits as well.That is 9 8 7 6 5 + 4 3 + 2 1 .
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You must keep the digits in order (from left to right).That is 9 + 8 7 + 6 5 + 4 3 2 1 is valid but 9 + 7 8 + 5 6 + 3 4 2 1 is not allowed.
The answer is 13.
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2 solutions
9+8+76+5-4+3+2+1= 100
3(6) + 4(1) = 22
and if we count + in front of 9 ( as we see in details and assumption) then also answer becomes 3(7) + 4(1) = 25
where is the mistake
Hey,u use num 5 twice? And it's wrong,right?
I read this article earlier and found the answer here .
how with 123-45-67+89= 100 ,A=1 , S=2 , then 3A+4S=3 1 + 4 2 = 11 but in your example, the first number include to A, but why in 98-76+54+3+21= 100 , A=3 not 4???
I have a better answer:123-45-67+89 ....answer of this ques. must be 11
Nice one! I got it in the second try though!
Here's my first attempt, 9 8 + 7 − 6 5 + 5 4 + 3 + 2 + 1 = 1 0 0
Bonus Question: Prove that this is the shortest method. xD
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We see that obviously we cannot have a 4 digit integer because e can have at most two four-digit integers and their difference would be over 4 4 4 4 , making it impossible to return to 1 0 0 .
Thus, we can work with at most three digits per integer. A quick check with a computer program will now easily prove that it is the one with the least number of signs.
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Oh yes sir! I just asked it simply. Thanks for your reply though. ⌣ ¨
- P.S: I guess it's morning up there, have a good day!
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@Sravanth C. – Hey , you were going to post some combinatorics problems.Why you are not posting? (Asked out of curiosity)
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@Nihar Mahajan – Yeah. My bad luck! I am out of station and I'm not able to post any questions, today though, I managed to write a few solutions. But I'll surely post 'em after I come back.
But don't let your curiosity down! keep it onnn. . . ;)
but here, the number 5 is used twice.
9 8 − 7 6 + 5 4 + 3 + 2 1
The least number of operators: 4.
3 × 3 + 1 × 4 = 1 3
Here is the list of all other possible combinations. All use more then 4 operators.
9 + 8 + 7 6 + 5 − 4 + 3 + 2 + 1
9 + 8 + 7 6 + 5 + 4 − 3 + 2 − 1
9 8 − 7 − 6 + 5 + 4 + 3 + 2 + 1
9 8 − 7 + 6 − 5 + 4 + 3 + 2 − 1
9 8 − 7 + 6 + 5 − 4 + 3 − 2 + 1
9 8 − 7 + 6 + 5 + 4 − 3 − 2 − 1
9 8 + 7 − 6 − 5 + 4 + 3 − 2 + 1
9 8 + 7 − 6 + 5 − 4 − 3 + 2 + 1
9 8 + 7 − 6 + 5 − 4 + 3 − 2 − 1
9 8 + 7 + 6 − 5 − 4 − 3 + 2 − 1
9 − 8 + 7 + 6 5 − 4 + 3 2 − 1
9 − 8 + 7 6 − 5 + 4 + 3 + 2 1
9 8 − 7 − 6 − 5 − 4 + 3 + 2 1
9 − 8 + 7 6 + 5 4 − 3 2 + 1
9 8 − 7 6 + 5 4 + 3 + 2 1 = 1 0 0