Sum of two adjacent numbers are primes: Difference between revisions
(Added C.) |
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num++ |
num++ |
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if num < 21 |
if num < 21 |
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? "n |
? "" + n + " + " + (n+1) + " = " + sum |
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else |
else |
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exit |
exit |
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<pre> |
<pre> |
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working... |
working... |
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1 + 2 = 3 |
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2 + 3 = 5 |
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3 + 4 = 7 |
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5 + 6 = 11 |
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6 + 7 = 13 |
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8 + 9 = 17 |
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9 + 10 = 19 |
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11 + 12 = 23 |
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14 + 15 = 29 |
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15 + 16 = 31 |
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18 + 19 = 37 |
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20 + 21 = 41 |
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21 + 22 = 43 |
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23 + 24 = 47 |
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26 + 27 = 53 |
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29 + 30 = 59 |
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30 + 31 = 61 |
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33 + 34 = 67 |
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35 + 36 = 71 |
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36 + 37 = 73 |
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done... |
done... |
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</pre> |
</pre> |
Revision as of 14:28, 21 January 2022
- Task
Show on this page the first 20 numbers and sum of two adjacent numbers which sum is prime.
C
<lang c>#include <stdio.h>
- define TRUE 1
- define FALSE 0
int isPrime(int n) {
if (n < 2) return FALSE; if (!(n%2)) return n == 2; if (!(n%3)) return n == 3; int d = 5; while (d*d <= n) { if (!(n%d)) return FALSE; d += 2; if (!(n%d)) return FALSE; d += 4; } return TRUE;
}
int main() {
int count = 0, n = 1; printf("The first 20 pairs of natural numbers whose sum is prime are:\n"); while (count < 20) { if (isPrime(2*n + 1)) { printf("%2d + %2d = %2d\n", n, n + 1, 2*n + 1); count++; } n++; } return 0;
}</lang>
- Output:
The first 20 pairs of natural numbers whose sum is prime are: 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Raku
<lang perl6>my @n-n1-triangular = map { $_, $_ + 1, $_ + ($_ + 1) }, ^Inf;
my @wanted = @n-n1-triangular.grep: *.[2].is-prime;
printf "%2d + %2d = %2d\n", |.list for @wanted.head(20);</lang>
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Ring
<lang ring> load "stdlibcore.ring" see "working..." + nl n = 0 num = 0
while true
n++ sum = 2*n+1 if isprime(sum) num++ if num < 21 ? "" + n + " + " + (n+1) + " = " + sum else exit ok ok
end
see "done..." + nl </lang>
- Output:
working... 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 done...
Wren
<lang ecmascript>import "./math" for Int import "./fmt" for Fmt
System.print("The first 20 pairs of natural numbers whose sum is prime are:") var count = 0 var n = 1 while (count < 20) {
if (Int.isPrime(2*n + 1)) { Fmt.print("$2d + $2d = $2d", n, n + 1, 2*n + 1) count = count + 1 } n = n + 1
}</lang>
- Output:
The first 20 pairs of natural numbers whose sum is prime are: 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
XPL0
<lang XPL0> include xpllib; int N, Num, Sum; [Text(0, "Working...^M^J"); N:= 0; Num:= 0; loop
[N:= N+1; Sum:= 2*N + 1; if IsPrime(Sum) then [Num:= Num+1; if Num < 21 then [Text(0,"N = "); IntOut(0,N); Text(0," Sum = "); IntOut(0,Sum); CrLf(0)] else quit ] ];
Text(0, "Done...^M^J"); ]</lang>
- Output:
Working... N = 1 Sum = 3 N = 2 Sum = 5 N = 3 Sum = 7 N = 5 Sum = 11 N = 6 Sum = 13 N = 8 Sum = 17 N = 9 Sum = 19 N = 11 Sum = 23 N = 14 Sum = 29 N = 15 Sum = 31 N = 18 Sum = 37 N = 20 Sum = 41 N = 21 Sum = 43 N = 23 Sum = 47 N = 26 Sum = 53 N = 29 Sum = 59 N = 30 Sum = 61 N = 33 Sum = 67 N = 35 Sum = 71 N = 36 Sum = 73 Done...