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Just out of curiosity, I wrote a maths equation today.

I then deleted one digit from the RHS and one digit from the LHS, but the equation still stood true.

I then again deleted one digit from the RHS and one digit from the LHS, but the equation still stood true.

Lastly, I deleted one digit from the remaining two digits on the RHS, even then after deleting a digit from LHS, the equation stood true.

Assuming, the equation only comprised of standard BODMAS operations(no exponents, logarithms etc.)

If all gaps left by erased digits were considered to be closed up,what were the three equations created?

EDIT:

There is no unique solution, but hardworked solutions will be appreciated.

Example:

3(43+2)=135

3(3+2)=15

3+2=5

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closed as too broad by Mithrandir, Glorfindel, boboquack, JonMark Perry, Sleafar Nov 26 '17 at 9:49

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Are the numbers used in the equation are formed by distinct digits (means no repeated digits) ?As, for example 4000 ÷ 4 = 1000 satisfies your requirements. Is it not so? $\endgroup$ – Mea Culpa Nay Nov 26 '17 at 7:29
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    $\begingroup$ Also, it may be clarified if the same digit gets deleted from each side. $\endgroup$ – Mea Culpa Nay Nov 26 '17 at 8:37
  • $\begingroup$ There is no compulsion to delete same digit from both sides $\endgroup$ – prog_SAHIL Nov 26 '17 at 9:12
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There seems to be very many solutions families, each having multiple solutions. Here is the simplest solution I found:

1111=1111 (at each phase, remove whichever digit)

Some of the other solution families:

N + 0 = N (remove corresponding digits of N).
N x 1 = N (same).
N + M = S, where N (mod 10) + M < 10, remove first digits on both sides.
N - M = D, similarly where N (mod 10) > M.

there are quite a lot of other solution families too.

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  • $\begingroup$ These are solutions but not worthy of accepting. $\endgroup$ – prog_SAHIL Nov 26 '17 at 9:13
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    $\begingroup$ So it’s a ”guess what I was thinking” puzzle then? Maybe you could try adding additional constraints to the puzzle so that only ”worthy” answers fulfilled all the requirements. $\endgroup$ – Bass Nov 26 '17 at 9:24
  • $\begingroup$ Absolutely not, but better answers are expected I will add an example to the ques. $\endgroup$ – prog_SAHIL Nov 26 '17 at 10:08
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    $\begingroup$ ”hardworked” is not a very good criterion, as it is entirely subjective, and falls smack in the middle of ”you are right, but keep guessing” territory of not-so-interesting puzzles. For example, I would claim that this answer is more hardworking than the example in the question. (The main reason being that the sample doesn’t follow the rules.) $\endgroup$ – Bass Nov 26 '17 at 10:42

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