ENIGMA 1538 — Six factors
by Richard England
Choose two 2-digit numbers, each of which is the product of two primes, and add them together
to produce a 2-digit result which is also the product of two primes.
All six primes are different.
Swap over the units digits between the two original numbers to create two new numbers,
each of which is the product of two primes. If we repeat the addition, once again all six primes are different.
In any sum
A + B = C
at least one of
A, B and C
must be even.
So therefore, one of the prime factors must be 2.
Let us first look at the possibility that
A is a multiple of 2.
(By symmetry, this also covers the case of
B being a multiple of 2.)
The other factor will be one of 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 and 43,
as 10 ≤ A < 90.
So possible values of A are 10, 14, 22, 26, 34, 38, 46, 58, 62, 74, 82 and 86.
Let us denote the values of A and B after swapping the units digits over as A' and B',
and look for values of B' which will have the same last digit as A,
and be chosen from the same set of possible values as A (but excluding the value of A itself).
This last digit of A and B' cannot be 0, as the two values must be different and there is no other candidate for A
that ends in 0.
If the last digit of A and B' is to be 2, the only possible values for A and B' are 22 and 62,
because the choice of B' = 82 would require 80 < B < 90, and thus A + B > 99.
We need to choose a last digit for B and A' such that each is a product of two primes, and so it the sum A + B.
Thus candidates for B are 61, 63, 65, 67 and 69 of which only 65 and 69 qualify as products of 2 different primes,
but the corresponding values for A' are 25 and 29, neither of which is a product of 2 different primes.
If the last digit of A and B' is to be 4, the only possible values for A and B' are 14 and 24, 14 and 74 or 24 and 74.
If B' = 24 candidates for B are 21, 23, 25, 27 and 29 of which only 21 qualifies as a product of 2 different primes,
but the corresponding values for A' is 11 which is not a product of 2 different primes.
If B' = 74 candidates for B are 71, 73, 75, 77 and 79 of which only 77 qualifies as a product of 2 different primes,
but the corresponding values for A' are 17 and 27 neither of which is a product of 2 different primes.
If the last digit of A and B' is to be 6, the only possible values for A and B' are 26 and 46.
Candidates for B are 41, 43, 45, 47 and 49 none of which qualifies as a product of 2 different primes.
If the last digit of A and B' is to be 8, the only possible values for A and B' are 38 and 58.
Candidates for B are 51, 53, 55, 57 and 59 of which only 55 qualifies as a product of 2 different primes.
The corresponding value for A' is 35 which is also a product of 2 different primes.
Thus we have a solution:
38 + 55 = 2 × 19 + 5 × 11
= 35 + 58 = 7 × 5 + 2 × 29 = 93 = 3 × 31
There is always the possibility of a solution in which
A + B is even.