In the year 4231, an astronaut travels to a planet with intelligent life. The astronaut saw numerous math equations written in the planet, and deduced that they were using a different set of notation. He saw the following arithmetic equations:
2
3
+
6
=
1
1
3
3
+
9
=
1
5
3
5
+
1
2
=
1
1
6
5
−
9
=
2
2
6
−
2
=
6
5
6
−
2
4
=
5
6
×
8
=
2
4
2
0
×
1
2
=
6
0
6
0
×
2
4
=
1
2
0
5
2
÷
4
2
=
2
6
9
÷
2
4
=
3
2
0
2
÷
1
0
1
=
1
0
1
∣
3
2
∣
=
6
∣
5
5
∣
=
∣
1
0
∣
=
1
∣
8
2
+
3
∣
=
∣
1
0
+
3
∣
=
∣
1
+
3
∣
=
4
∣
3
6
−
2
3
∣
=
∣
9
−
5
∣
=
4
∣
6
×
8
∣
=
2
4
∣
2
0
×
1
2
∣
=
6
0
∣
5
2
÷
4
2
∣
=
2
∣
6
9
÷
2
4
∣
=
3
In order for him to gain an audience with the ruler, he needs to solve this equation using these same rules:
∣ 3 2 ÷ { 3 6 − [ 5 8 + ( 3 2 × 3 ) ] × [ 4 5 + ( 3 6 − 2 ) ] } ∣ .
Help him to solve this strange problem.
Hint
: To find logic in these equations, you can use LCM, GCD, and the sum of the digits of these numbers.
The order of operations is the same on this planet and on Earth: First the parentheses, after the brackets, braces, multiplication, division, etc.
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Exactly!
What do you mean by keep performing? Also |6×8|=24 and that's not a single digit.
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He forget to say it. The absolute value of a multiplication or division not change. But in addition or subtraction have a change: you should add all digits of the number until get a single number and make a normal addition or subtraction.
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Yeah, sorry about forgetting that. Thanks for the fun problem Victor!
Thanks! Also, if I'm not mistaken, the first three clues seem like that the + symbol could mean the first value substracted by the double of the second value. For example
2 3 + 6 = 2 3 − 1 2 = 1 1 3 3 + 9 = 3 3 − 1 8 = 1 5
So I stuck with that, and didn't get the answer...
It would have been better if the hint wasn't given
I wonder why |32| = 5 and |36-23| = |9-5| = 3
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I made a mistake. |36-23| = |9-5| = 4.I already solved this mistake. Sorry.
And how is |9-5 |= 3 and|3+1| = 4 ?
the most complicated question
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The logic of this planet is as follows:
For a + b , take the sum of digits of a and the sum of digits of b and add them together.
For a − b , do the same as 1 except subtract the sums.
a × b = LCM of a , b
a / b = GCF of a , b
∣ a ∗ b ∣ where * is an operation means keep performing until the result is a single digit.