## Task 36/75

Choose any two-digit prime number with a cross sum of 10 and subtract the number 18 from it as many times as necessary until the difference is between 10 and 20. Quadruple this difference! Place the difference in front of this product!

How many "starting numbers" are there for the calculation, and why is the result unique?
Two-digit numbers with a cross sum of 10 are the numbers 19; 28; 37; 46; 55; 64; 73; 82; 91. Among these, only three are prime numbers: $19 ; 37 ; 73$.

Since the prime number is already between 10 and 20, according to the instructions of the problem, the number 18 is to be subtracted zero times. For the other two starting values, subtracting the number 18 once or three times leads to the same intermediate result of 19. Starting from this uniquely determined intermediate result, the final result is the number 1976.

Answer to the additional question: There are thus three possible starting numbers for the calculation. The result is still unique because all three numbers have the same congruence mod 19:

$$
10 \equiv 1 \equiv 37 \equiv 1 \equiv 73 \equiv 1(\bmod 18)
$$

and because there is exactly one number between 11 and 20 with this congruence mod 18.

### 2.16 Problems and Solutions 1976