Re: Enigma 1539 - Board with numbers

johnjo wrote:
On May 20, 7:41 am, Chappy <petergregorychap...@xxxxxxxxxxx> wrote:
Enigma 1539 - Board with numbers
New Scientist magazine, 4 April 2009.
By Susan Denham.

As I have mentioned in a previous Enigma
puzzle, a "Snakes and Ladders" board
consists of a 10-by-10 grid of squares. In
row 1 (at the bottom) the squares are
numbered from 1 to 10 from left to right.
Then in row 2 the squares are numbered 11 to
20 from right to left. In row 3 they are 21
to 30 from left to right again, and so on,
ending up at square 100 in column 1 and row
10.

I have been cutting up such a board. I have
cut out a rectangle consisting precisely of
some of its squares. It runs from a prime-
numbered row of the original board to a
higher prime-numbered row inclusive, and it
runs from a prime-numbered column to a higher
prime-numbered column inclusive. Furthermore,
the total of all the number in my rectangle
is a prime.

What is that total?

Ciao,
Chappy.

How can this be deduced except by trial and error?
A spreadsheet suggests 449 is the only such prime sum.
being row 3:7 column 2:3.
HTH
JJ

Let the low numbered row be R, and number of columns w, and the number of rows h.

Since successive rows have numbers in opposite orders, the sum of pairs of numbers in consecutive rows is a constant for a given pair of rows - that is, it doesn't change with column number. So if the number of rows is even, we can let h' = h, and write the total as

[(2R + h' - 2) * 10 + 1] * w * h' / 2. [A]

The result will be divisible by w, which cannot itself be one. So if the number of rows is even, then the total is not prime.

So the number of rows is odd. The total is therefore the sum of the above expression [A], for some R and with h' = h - 1, and the sum of the columns in a remaining row r. We can choose r so that it's odd (the row being either above the pairs of rows, or below it). In an odd numbered row, the numbers are ascending. If the starting column is c, then the sum of the numbers used in the row is

(r - 1) * 10 * w + (2c + w - 1) * w / 2 [B]

If w is odd, then both this expression and the expression [A] will be divisible by w, and the total will not be prime.

If w is even, but not prime, then the total will be divisible by w/2. So w is 2. The only way to get w to be 2 is for c to be 2 - that is columns 2 and 3 are used.

Further, since h is odd, consideration of the possible pairs of rows tells us that the starting row number must be 3 or 5, and the number of rows, h, must also be 3 or 5. Since the row number is odd, R = r + 1.

Substituting for b and c in [B] reduces it to

(r - 1) * 20 + 5 [B]

Substituting for R, h' and w in [A] gives

[(2r + h - 1) * 10 + 1] * (h - 1) [A]

If h is 3, then considered modulo 3,

[B] becomes (r - 1) * 2 + 2 mod 3 = 2r

[A] becomes (2r - 1) * 1 + 1) * (-1) mod 3 = -2r mod 3

and the total is 0 mod 3. That is, if h is 3 then the total is divisible by 3.

That only leaves h = 5, for which r must be 3.

[B] becomes 45

[A] becomes 404

The total is 449, which is prime.

Sylvia.

















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