Puzzle - RPL and others

[Albert Chan]

(04-23-2021 04:08 PM)Albert Chan Wrote:  This filled all even numbers

Code:

1  2  3  4  5  6  7  8  9
=========================
   4     2  5  8     6
   8     6  5  4     2

I noticed odd digits [1,3,7,9], mod7 is [1,3,0,2], or simply 0 to 3.
This may simplify mod7 calculations (without calculator)

For example, lets try to solve missing digits of last pattern.
Let a = 1st digit, b = 3rd digit, c = 7th digit

10^6*a + 10^4*b + c + 806540 ≡ a - 3*b + c ≡ 0 (mod 7)

a+c ≡ 3*b (mod 7)

a+c ≡ 3*0 ≡ 0, no solution
a+c ≡ 3*1 ≡ 3, [a,c] ≡ [0,3] or [3,0]
a+c ≡ 3*2 ≡ 6, no solution
a+c ≡ 3*3 ≡ 2, [a,c] ≡ [0,2] or [2,0]

Passes mod7 test: [a,b,c] = [7,1,3], [3,1,7], [9,3,7], [7,3,9]

Digit(6,7,8) for divisible by 8: 492 (mod 8) ≠ 0, last case rejected.
Digit(1,2,3) for divisible by 3: 781, 381, 983, only 381 passes.

-> soc-sec# = 381 654 729