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Any insights are welcome for this puzzle.

The following equation is wrong: $103 - 102 = 3$. Move one numeral to make it correct. The numeral moved is: $0,1,2$ or $3$?

S.D.
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  • Can you move the same number from both $102$ and $103$?If you can, you get the equation $13-12=(3^0)^0$ – Konstantinos Gaitanas May 06 '14 at 07:12
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    It's not a trick solution is it? - move one of the $1$s on top of the equals sign to make it look like $\ne$? – David May 06 '14 at 07:16
  • @David I have always tried such a solution in similar puzzles but it was never correct...I think it should be an equation which will be correct. – Konstantinos Gaitanas May 06 '14 at 07:20
  • @KonstantinosGaitanas it was the wording that made me suspicious - it doesn't actually say "move one numeral to make a correct equation". . . – David May 06 '14 at 07:22
  • @KonstantinosGaitanas: All conditions I know of are in the formulation, so I have no idea whether it shall be just a single move or we can move the same numeral. I think, with a single one nothing works so anyways it is worth trying with two. I was not sure whether additional operations are permitted though: that is, can we find a solution of the form $x - y = z$ with no more operations? – S.D. May 06 '14 at 07:23
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    @David I've had the same puzzle but it didn't specify "numeral" and there was no follow-up question. Naturally I changed it to 103102 ≠ 3 – OrangeDog May 06 '14 at 10:41

4 Answers4

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Konstantinos was on the right track.

$$103 - 10^2 = 3$$

Since the question was "how to solve it", for any interested...

I solved it by first seeing that all the obvious a + b = c can be eliminated by seeing the range of the reduced or increased numbers. Konst suggested exponents, which is the only other operator I can think of, $1$ is too small to be helpful, but 2 or 3 could work.

DanielV
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    I thought it would be the 2. To say that the numeral was 0, 1, or 3 would have been vague because there were two copies of each. – JRN May 06 '14 at 07:24
  • Thanks, but the same question here as to @Konstantinos: I am not sure whether additional operations are allowed. Is there a solution of the form $x - y = z$ where $x,y,z$ are just simple digital expressions ($102$ is simple, $10^2$ is not)? – S.D. May 06 '14 at 07:25
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    @S.D. No, you can see that by reducing any number from 103 you get a negative difference, reducing any number from 102 gets a difference in excess of 80 (which the 3 can't create), and removal from 3 doesn't leave any valid grammatical expression behind. Therefore there must be some interpretation necessary. – DanielV May 06 '14 at 07:30
  • @DanielV: thanks, I had similar arguments - but just thought I overlooked some thing. – S.D. May 06 '14 at 07:32
2

Alternative solution: use Roman numerals:

CIII - CII = III
         ^
         move this one

CIIII - CI = III
    ^
    to here

(or

CIII - CII = III
               ^
               move this one

CIIII - CII = II
    ^
    to here

)

Hugh Bothwell
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0

Just move the digit $2$ upward to the superscript position, so that it appears as an exponent of $10$ without adding an operator symbol.

RobertK
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-1

The solution indeed uses exponents:

$103-10^2=3$

user148285
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