easy and uneasy riddles

On 15 January, The Riddler had both a straightforward and a challenging riddles. The first one was to optimise the choice of a real number d with the utility function U(d,θ)=d ℑ(θ>d), when θ is Uniform(0,100). Leading unsurprisingly to d=50…

The tough(er) one was to solve a form of sudoku where the 24 entries of a 8×3 table are integers in {1,…,9} and the information is provided by the row-wise and column-wise products of these integers. The vertical margins are

294, 216, 135, 98, 112, 84, 245, 40

and the horizontal margins are

8 890 560, 156 800, 55 566

After an unsuccessful brute-force (and pseudo-annealed) attempt achieving a minimum error of 127, although using the prime factor decompositions of these 11 margins, I realised that some entries were known: e.g., 7 at (1,2), 5 at (3,2), and 7 at (7,3), and (much later) that the (huge) product value for the first column implied that each term in that column had to be the maximal possible value for the corresponding rows, except for 5 on row 7. This leads to the starting grid

    [,1] [,2] [,3]
[1,]    7    7    6
[2,]    9    0    0
[3,]    9    5    3
[4,]    7    0    0
[5,]    8    0    0
[6,]    7    0    0
[7,]    5    7    7
[8,]    8    0    0

and an additional and obvious exclusion based on the absence of 3’s in the second column, of 5’s and 2’s in the third column shows there was a unique solution

    [,1] [,2] [,3]
[1,]    7    7    6
[2,]    9    8    3
[3,]    9    5    3
[4,]    7    2    7
[5,]    8    2    7
[6,]    7    4    3
[7,]    5    7    7
[8,]    8    5    1

as also demonstrated by a complete exploration with R:

Try it online!

le compte est bon

The Riddler asks how to derive 24 from (1,2,3,8), with each number appearing once and all operations (x,+,/,-,^) allowed. This reminded me of a very old TV show on French TV, called Le compte est bon!, where players were given 5 or 6 numbers and supposed to find a given total within 60 ,seconds. Unsurprisingly there is an online solver for this game, as shown above, e.g., 24=(8+3+1)x2. But it proves unable to solve the puzzle when the input is 24 and (2,3,3,4), only using 2,3 and 4, since 24=2x3x4. Introducing powers as well, since exponentiation is allowed, leads to two solutions, (4-2)³x3=(4/2)³x3=(3²-3)x4=3/(2/4)³=24… Not fun!

I however rewrote an R code to check whether 24 was indeed a possibility allowed with such combinations but could not find an easy way to identify which combination was used, although a pedestrian version eventually worked! And exhibited the slightly less predictable 4^3/2x3=24!