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Here is a simple math problem. Can you solve it?

Examples:

  • 678+57-24*2 = 7
  • 265+34-12*3 = 8
  • 328+58-22*1 = 12

Problem:

541+13-21*0 = ?

HINT 1:

Will see if this hint makes any sense or will edit it in another way, sorry making hints are hard for this(atleast for me):P

678+ 57- 24* 2 = 7

HINT 2:

(678+)(57-)(24*)(2) = 7

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  • $\begingroup$ @I am the Most Stupid Person, Will add the Hint tomorrow till then you can try, I am sure someone will find the solution might be you:P $\endgroup$
    – Preet
    Dec 14, 2017 at 7:23
  • $\begingroup$ @Preet Is the answer is 21? $\endgroup$
    – rudra
    Dec 14, 2017 at 14:07
  • $\begingroup$ @rudra i think no but you can add a answer, you might be on right track $\endgroup$
    – Preet
    Dec 15, 2017 at 0:10
  • $\begingroup$ Anyone please explain the reason for down vote..? Is it because of hint or something else? $\endgroup$
    – Preet
    Dec 15, 2017 at 0:57
  • $\begingroup$ The first hint was kind of misleading. I thought we were supposed to fill in the blanks. $\endgroup$
    – ibrahim mahrir
    Dec 15, 2017 at 1:31

1 Answer 1

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I think I found if, but it is a bit complicated.

The hint(added spaces) seems to imply that the symbols affect the previous number so.
678+ 57- 24* 2 = 7
21(6+7+8) 2(7-5) 8(2*4) 2
21 + 2 - 8 * 2 = 7 (Then I find a chain of formula with the resulting numbers that gives to proper answer for all 3 examples)

265+34-12*3 = 8
13 1 2 3
13 + 1 - 2 * 3 = 8

328+58-22*1 = 12
13 3 4 1
13 + 3 - 4 * 1 = 12

Now if I apply that chain of equation(+-*) to the problem, I get
541+13-21*0 = ?
10 2 2 0
10 + 2 - 2 * 0 = 12

Note that the final chain of equations is the same as the initial one, so basically we need to apply each equations 2 times, once within the number itself and then between each numbers.

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  • $\begingroup$ You got it!! was just adding another hint:P that means my previous hint was correct:P. $\endgroup$
    – Preet
    Dec 15, 2017 at 1:28
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    $\begingroup$ @Preet A variation of the Reverse Polish notation. Nice one! Good job @ stackReader! $\endgroup$
    – ibrahim mahrir
    Dec 15, 2017 at 1:30

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