# A riddle that has been killing me the whole day

So I'm walking around in London and found the following number riddle. The rules say, that what ever pattern you find, must be true for the rows as well as the columns. The answer in level 1 is for example 33 since it follows a+b+2=c. Meaning the third block always is the sum of the two first ones plus two, both in the columns and the rows.

0

The formula is:

c = a²+(b-10)*a

Examples:

16 = 4² +(10-10)*4,
-6 = 1² + (3-10)*1,
0 = (-20)² + (30-10)*(-20),
30 = 10²-(3-10)*10,
-20=4²+(1-10)*4

• Awesome, thanks a lot! Now I can sleep again Oct 25, 2018 at 14:06
• Out of interest, hos did you find the solution Oct 25, 2018 at 14:07
• It was like the eigth or ninth idea I had. The 16 looked suspiciously like 4², so I just had to cancel the 10. I also had the gut feeling to square the 10. The 10, 3, 30 was quite a help, as it invalidated lots of my earlier ideas. The 4, -1, -20 was a large red herring to me, as I tried to get to 4*(-5) or -4*5 or -25+5 or things like that a lot. I could have walked you through my thought process a bit better, but I threw away my notes. I guess you could sum it up as a combination of luck, intuition, patience and more luck. Oct 29, 2018 at 9:27

90

Step-by-step breakdown for all puzzles.

Following the nomenclature (Columns a,b,c from left to right respectively):

1.

c=a+b+2
i.e. c = 16+15+2 = 33

2.

c = b * a/4
i.e. c = 5 *16/4 = 20

3.

c = a^2 - b^2
i.e. c = 12^2 - (-5)^2 = 119

4.

c = 19 - a - b
Funfact: all rows & columns sum up to 19 i.e. c = 19 - (-3) - 16 = 6

5.

c = (b-a)^2 - sqrt(a)*10
But this should not be correct, unless u take c = 2500 -10*sqrt(20)i as an answer ....
Hence, I explored methods
c = 5[mod(b,a) * (b/a)-1] - a
i.e. c = 5[mod(30,-20) * (30/-20)-1] -(-20) = 90

• I got the same in all answers but not level five seems uncorrect (which I still have not solved). The answer from the column is real number, and from the row is complex, and they dont agree Oct 17, 2018 at 7:14
• Edited. See if these operators are allowed :p Oct 17, 2018 at 9:13
• The new method you describe, doesnt Seem to work with 4,1 Since it gives -9 and not -20 Oct 19, 2018 at 22:41

The fifth equation is NOT of the form

axy + bx + cy + d = z

For any set of constants a, b, c, d.

Because

Credit to https://matrixcalc.org

• Nice observation, but one could just calculate the determinant of the matrix on the left hand side, and conclude, that there is not a unique solution of the desired form. Oct 17, 2018 at 12:18

If there's no solution for axy + bx + cy + d = z, you could consider the fact that horizontal and vertical both equal 10 when added together with no transformations.

Using that:

a + b != c
30 + (-20) != c, 16 + (-6) != c
10 != c, 10 != c
{c ∈ ℝ, c != 10}

• I don't think my partial answer necessarily implies at ax + by != z. It requires that there be additional (higher order or more complex) terms, however, there may be some values of x and y for which those terms cancel. Oct 23, 2018 at 0:34

12.

$$-20+30+2=12$$, and

$$16-6+2=12$$

Then:

$$-20+30+2 = 12$$

Or did I miss something?

• Welcome to Puzzling.SE! The rule has to work for all rows and columns, so unfortunately this would not work for the first column: 4 + 1 + 2 != -20
– obl
Oct 19, 2018 at 22:11