Description

COMP9021, Session 1, 2015
1 R Finding particular sequences of prime numbers
Write a program that finds all sequences of 6 consecutive prime 5-digit numbers, so of the form (a,b,c,d,e,f) with b = a + 2, c = b + 4, d = c + 6, e = d + 8, and f = e + 10. So a, b, c, d and e are all 5-digit prime numbers and no number between a and b, between b and c, between c and d, between d and e, and between e and f is prime.
The expected output is:
The solutions are:
13901 13903 13907 13913 13921 13931
21557 21559 21563 21569 21577 21587
28277 28279 28283 28289 28297 28307
55661 55663 55667 55673 55681 55691
68897 68899 68903 68909 68917 68927
2 R Decoding a multiplication
Write a program that decodes all multiplications of the form
* * *x * *
———-
* * * *
* * *
———-
* * * *
such that the sum of all digits in all 4 columns is constant.
The expected output is:
411 * 13 = 5343, all columns adding up to 10.
425 * 23 = 9775, all columns adding up to 18.
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3 Decoding a sequence of operations
Write a program that finds all possible ways of inserting + and – signs in the sequence 123456789 (at most one sign before any digit) such that the resulting arithmetic expression evaluates to 100. Here are a few hints.
• 1 can either be preceded by -, or optionally be preceded by +; so 1 starts a negative or a positive number.
• All other digits can be preceded by – and start a new number to be subtracted to the running sum, or be preceded by + and start a new number to be added to the running sum, or not be preceded by any sign and be part of a number which it is not the leftmost digit of. That gives 38 possibilities for all digits from 2 to 9. We can generate a number N in [0,38 − 1]. Then we can:
– consider the remainder division of N by 3 to decide which of the three possibilities applies to 2;
– consider the remainder division of N3 by 3 to decide which of the three possibilities applies to 3;
– consider the remainder division of 3N2 by 3 to decide which of the three possibilities applies to 4; – …
The expected output is (the ordering could be different):
1 + 23 – 4 + 5 + 6 + 78 – 9 = 100
123 – 4 – 5 – 6 – 7 + 8 – 9 = 100
123 + 45 – 67 + 8 – 9 = 100
123 + 4 – 5 + 67 – 89 = 100
12 + 3 + 4 + 5 – 6 – 7 + 89 = 100
123 – 45 – 67 + 89 = 100
12 – 3 – 4 + 5 – 6 + 7 + 89 = 100
1 + 2 + 34 – 5 + 67 – 8 + 9 = 100
1 + 2 + 3 – 4 + 5 + 6 + 78 + 9 = 100
-1 + 2 – 3 + 4 + 5 + 6 + 78 + 9 = 100
12 + 3 – 4 + 5 + 67 + 8 + 9 = 100
1 + 23 – 4 + 56 + 7 + 8 + 9 = 100
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