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$62-5+3*2=25$

Move a number's position in order to get the right equation.

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  • $\begingroup$ We can just move a digit or a number but not add any other mathematical function? right? $\endgroup$ – Netham Sep 12 '17 at 13:30
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    $\begingroup$ @Netham yes, only the number. eg., if you're moving the "-5" somewhere, you should move it as a whole including the operator that precedes it, like that "-5." $\endgroup$ – jdoe179 Sep 12 '17 at 14:07
  • $\begingroup$ ^ "move only the number"..."move operators too" $\endgroup$ – feelinferrety Sep 12 '17 at 15:41
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    $\begingroup$ @jdoe179 Can you please update the question and fix the missing part pointed by @ feelinferrety $\endgroup$ – Laschet Jain Sep 13 '17 at 21:09
  • $\begingroup$ If this is move one number with one operator I don't think it can be solved. Please clarify. $\endgroup$ – paparazzo Sep 14 '17 at 14:31
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Can't tell if moving one digit is all we're are allowed, but this one is SOOO close:

$6^2-5-3\times2=36-11=25$.

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    $\begingroup$ how did the 3x2 become negative? $\endgroup$ – Jason V Sep 13 '17 at 15:41
  • $\begingroup$ @JasonV 6^2-5-3×2 =36-5-6 = 31-6 = 25. It seems correct or did I missed something? $\endgroup$ – Alex Sep 14 '17 at 19:09
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    $\begingroup$ @Alex that is a correct statement, but the original equation has 62 - 5 + 3x2.... this answer uses 62 -5 *-*3x2. $\endgroup$ – Jason V Sep 15 '17 at 13:38
  • $\begingroup$ Given that addition having higher precedence than subtraction (and multiplication than division) is somewhat common misconception due to the order of the letters in PEMDAS, I wouldn't be too surprised if this is the intended answer, if the OP had intended 62-5+3*2 to be parsed as 62-(5+3*2). $\endgroup$ – hvd Sep 15 '17 at 15:40
  • $\begingroup$ @hvd the only way that happens is with the parentheses. $\endgroup$ – Jason V Sep 15 '17 at 21:24
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62 is equivalent to 43+19. so we have 43+19-5+3x2=25. By moving the +19 to the RHS we are left with

The answer

43-5+3x2=25+19 ==> 43-5+6 = 44 ==>44=44

Reasoning

I began this approach by looking into re-writing the constants as quotients (as in 275/55=5, then move the numerator 275 above the 25 to get 11, but this equation eventually works out to 13=11...) I am aware that my final method is not as eloquent as my original attempt, but I think it warrants being a correct answer. (and possibly spark some new ideas outside of the box) Afterall, I don't think anyone would argue with (62 is equivalent to 43+19).

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  • $\begingroup$ moving +19 to RHS would make it -19. $\endgroup$ – Abhishek Sep 14 '17 at 11:51
  • $\begingroup$ @Abhishek that is the whole purpose of the puzzle, to not use the rules of math, but to simply move one operand and its leading operator. $\endgroup$ – Jason V Sep 14 '17 at 12:49
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I think, following is right position.

25 - 6 + 3 x 2 = 25

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    $\begingroup$ That looks to me like more than one number has to be moved. What am I missing? $\endgroup$ – Gareth McCaughan♦ Sep 12 '17 at 13:53
  • $\begingroup$ Yep, looks like a number swap instead of moving one number only. $\endgroup$ – jdoe179 Sep 12 '17 at 13:54
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    $\begingroup$ @GarethMcCaughan I had the same doubt. It seems to assume that moving the number would keep the operations in the same indexes. i.e. 6253225 becomes 2563225 by moving the 6, then replacing the operations in their original indexes gives this answer. $\endgroup$ – Apep Sep 12 '17 at 13:58
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    $\begingroup$ Not even a number swap. It's a cyclic permutation of the first three digits. $\endgroup$ – Gareth McCaughan♦ Sep 12 '17 at 13:58
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Can you do:

62 - 5 + m * 2 = 25 (assuming I can move the position of 3 so its on its side and then count it as an m)
simplified is then:
57 + 2m = 25
2m = -32
m = -16
It's an equation. if its the right one is up to preference

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Using the updated rule that you move an attached operator with the digit you can just move the

x2

to the other side to get

62-5+3=25x2 ... (both sides now equal 50)

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  • $\begingroup$ That math doesn't work. Even if you fix the numbers (you flipped a 3 and a 5), it doesn't work. $\endgroup$ – Rubio♦ Sep 13 '17 at 23:33
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    $\begingroup$ Ah yes. 50 doesn't quite equal 60... $\endgroup$ – Penguino Sep 14 '17 at 2:47

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