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e pluribus unum ii (Posted on 2004-01-22) Difficulty: 3 of 5
Below are three groups of three numbers each. Combine the numbers in each group using the standard binary operations (addition, subtraction, multiplication, division, and exponentiation) so that each group yields the same number.
  1. 3, 15, 18
  2. 10, 13, 36
  3. 24, 27, 39

For an example, see the first problem.

See The Solution Submitted by DJ    
Rating: 4.6667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: an error in your DATA | Comment 5 of 12 |
(In reply to an error in your DATA by Ady TZIDON)

I've changed that 5 to a 15 and the results still do not have a triple:


-12 2 2 36 / (10 - 13)
-12 3 2 24 / (27 - 29)
-13 1 1 (15 / 3) - 18
-13 2 1 (10 - 36) + 13
-39 1 2 15 - (18 * 3)
-39 2 1 (10 - 13) - 36
-8.333333333333333D-02 2 1 (10 - 13) / 36
-8.333333333333333D-02 3 1 (27 - 29) / 24
12 2 2 36 / (13 - 10)
12 3 2 24 / (29 - 27)
13 1 2 18 - (15 / 3)
13 2 1 (36 - 10) - 13
39 1 1 (18 * 3) - 15
39 2 1 (13 - 10) + 36
8.333333333333333D-02 2 1 (13 - 10) / 36
8.333333333333333D-02 3 1 (29 - 27) / 24

----
Exponentiation is considered as in
CASE 5
IF LOG(opnd(eq, 1)) * opnd(eq, 2) < 300 * LOG(10) THEN
rslt1a = opnd(eq, 1) ^ opnd(eq, 2)
ELSE
rslt1a = 1: err1 = 1
END IF

It is the case that when the result is huge, above about 10^300, we ignore the answer,
but unmatched lines in the original output do show up as in:
14348925 1 1 (3 ^ 15) + 18
It's just that none of these match anything.
  Posted by Charlie on 2004-01-22 13:19:05
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