by Victor Bryant
published Friday April 12 2024
I started with a six-by-six grid with a 0 or 1 in each of its 36 squares: they were placed in chequerboard style with odd rows 101010 and even rows 010101. Then I swapped over two of the digits that were vertically adjacent. Then in three places I swapped a pair of horizontally adjacent digits.
In the resulting grid I read each of the six rows as a binary number (sometimes with leading zeros) and I found that three of them were primes and the other three were the product of two different primes. The six numbers were all different and were in decreasing order from the top row to the bottom.
What (in decimal form) were the six decreasing numbers?
Sunday Times Teasers Discussion Group 
Because it isn’t excluded we have to assume a digit could be involved in 2 or even 3 swaps vertically ?
A combined analysis and synthesis approach led to a short route to the answer. Only at one point was there more than option to consider, and there was only one way forward. I was amused by the difference between the initial and final sum of the six rows.
Yes there is a difference, but just A difference!
should read “horizontally’. However. swapping a digit a second time does not change the number and swapping 3 times disallows 6 different numbers. It seems 2 swaps of different pairs in the same line can still allow 6 different numbers.
The bits have to be adjacent so there are five different possible results for a row.
…and indeed that a digit swapped vertically could later be swapped horizontally.
..but the second swap would not change the number ?
All resulting binary numbers must end in a 1 as all are odd numbers (except 000010 which is 2 in decimal, a prime but to get this involves 2 vertical swaps ?)
so actually one number has to be even (the “product” with one multiplier of value 2) . Note, multiplying a binary by 2 shifts all digits to the left.
One or more numbers have to be even.
There is nothing stopping more than one of the products incuding a 2.
What an enjoyable little teaser. I’m on holiday and took little paper with me, this one solved on 3/4 of a side of A4.
I wondered at first about swaps “colliding” but when I considered what wouldn’t change if this were the case all became clear. One of the 6 numbers was quickly determined.
Then I considered the vertical swap giving two vertically adjacent numbers, these two of the 6 numbers were then determined.
Then I considered the horizontal swaps. Knowing the starting points fixed one of the 3 remaining numbers and the positions of all of them, allowing the 2 final numbers to be determined as the only ones that go in the remaining gaps in the sequence. This also verified the answer in unique. Nice.
Can’t be less cryptic this early on in the week.
At least 5 numbers must change. At most 5 numbers can change.
Indeed. The only number that can stay intact is one of the 21s, so there must be three A and one V (affecting two rows), and all on separate rows.
Dealing with As first, the only 21 that can stay intact is R6, as there need to be two 21-original Rows above higher than that.
At last, a teaser is presented that I was not tempted to resort to high level language
programming or use of a spreadsheet (and often failing to solve).
Although the teaser does not stipulate that the two adjacent horizontal rows involved
in the vertical swap are excluded in the subsequent horizontal swaps, or that more than
one swap is not allowed in a single row, this must be the case. Otherwise there
will be at least two rows untouched by the swapping processes, one of which would
violate the criterion that the 6 numbers are either prime or have only two prime factors.
A ‘010101’ row, equivalent to 21=3×7, satisfing the two prime factor criterion is o.k.
but a second (‘101010’ = 2x3x7) cannot. It then follows that, in accord with the
comment above, 5 rows are changed and one is not
Solution then reduces to juggling the 13 possibilities: (i) 6 from the vertically adjacent digit swap
in two adjacent rows; (ii) 7 possiblilies of horizontal swaps in ‘101010’ and ‘010101’ rows.
Included in this juggling process is deciding where in the 6×6 grid the two rows involved in the first
swap operation, consistent with the final 6 numbers be in descending order.
I believe I have a unique answer.
Another nice teaser – but not yet sure whether I have demonstrably shown uniqueness.
Regarding decimal sums of rows initially and finally, can also note that these sums and their sum each have two different prime factors and all six are different. Furthermore, five of these are used in factorising the last four numbers in this teaser’s answer. Numbers have a life of their own?
The 12th prime is 37.
3×7=7×3=21 (or 10101).
The 21st prime is 73 (or 1001001).