## Day 280

### October 9, 2017

Pat Ballew is a retired math teacher who writes a blog On This Day In Math that gives a day-by-day history of mathematics. The blog is odd, quirky, and unquestionably fun. On October 7th, Ballew wrote:

The 280th day of the year…. The sum of the first 280 consecutive primes, mod 280, is prime.

Since I like to play with prime numbers, that got my attention, and I quickly went to determine how many such days there are in a year.

Your task is to determine how many days in a year share the characteristic that on the nth day the sum of the first n primes, mod n, is prime. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Pages: 1 2

### 8 Responses to “Day 280”

1. Using the primes package for Haskell, the simple solution works just fine:

```module Main where

import Data.Numbers.Primes (isPrime, primes)

sumPrimesMod :: Int -> Bool
sumPrimesMod n = isPrime (sum (take n primes) `mod` n)

main :: IO ()
main = do
let ns = filter sumPrimesMod [1 .. 367]
print ns
print \$ length ns
```

I went up to 367 to include leap years.

2. Perl ‘golf’ish solution – using a bit of bash and a bit of perl!

```perl -e '%p=map{\$_,1}@ARGV;print"@{[grep{\$p{(\$t+=shift)%\$_}}1..366]}\n"' `seq 2 2473|factor|sed 's/.*: //g;/ /d'| xargs`
```

The bash in back ticks generates a list of 365 primes…
Get a list of 108 numbers – can add “|wc -w” to the end to get the number of values which satisfy this….

3. mcmillhj said

SML:

```fun isPrime n =
if n = 2 then true
else if n < 2 orelse n mod 2 = 0 then false
else let
fun loop k =
if k * k > n then true
else if n mod k = 0 then false
else loop (k + 2)
in loop 3
end ;

fun primesUpTo n = let
fun loop n 0 acc = rev acc
| loop n count acc = if isPrime n
then loop (n + 2) (count - 1) (n :: acc)
else loop (n + 2) count acc

in
loop 3 (n-1) 
end ;

val primes = primesUpTo 365;
fun sumPrimesMod n = let
val nPrimes = List.take(primes, n)
val sum = foldl (op +) 0
in
isPrime (sum nPrimes mod n)
end ;

val matchingDays = List.filter (fn x => sumPrimesMod x) (List.tabulate(365, fn x => x + 1));
(* [5, 6, 7, 8, 12, 15, 16, 19, 20, 21, 24, 26, 30, 34, 37, 38, 40, 42, 44, 45,
46, 48, 49, 50, 55, 58, 59, 60, 62, 64, 65, 66, 67, 68, 70, 72, 73, 75, 76,
78, 86, 87, 88, 92, 102, 116, 120, 122, 124, 128, 130, 132, 135, 140, 143,
145, 150, 156, 158, 164, 165, 166, 168, 172, 173, 175, 176, 182, 183, 191,
196, 210, 214, 216, 218, 223, 234, 236, 241, 248, 250, 256, 259, 262, 265,
266, 272, 280, 285, 301, 306, 310, 311, 314, 315, 324, 328, 330, 336, 337,
344, 347, 348, 349, 352, 355, 358, 365] *)
```
4. Rutger said
```def is_prime(n):
return (n > 1) and (n == 2 or n % 2) and all(n % x for x in range(3, int(1 + n**0.5), 2))

def first_n_primes(n):
result, i, x = [], 0, 2
while i < n:
if is_prime(x):
result.append(x)
i += 1
x += 1
return result

first_365 = first_n_primes(365)

for day in range(1,365):
if is_prime(sum(first_365[:day]) % day):
print(day)

```
5. matthew said

Here’s another Haskell solution, this one builds a list of the partial sums with scanl rather than recomputing each time around:

```import Data.Numbers.Primes (isPrime, primes)
main = print \$
map snd \$
filter (isPrime . uncurry mod) \$
take 366 \$
zip (scanl1 (+) primes) [1..]
```
6. chaw said

Using standard (R7RS) Scheme and popular a couple of popular libraries
(SRFI-1 and SLIB’s factor):

``` (import (scheme base) (scheme write) (only (slib factor) primes>) (only (srfi 1) iota fold filter take-while)) (display (let* ((n 366) (nprimes (primes> 1 n)) (sum-nprimes (reverse (fold (lambda (p sums) (cons (+ p (car sums)) sums)) (list (car nprimes)) (cdr nprimes)))) (mprimes (take-while (lambda (v) (< v n)) nprimes))) (length (filter (lambda (pr) (member (modulo (cdr pr) (car pr)) mprimes)) (map cons (iota n 1) sum-nprimes))))) (newline) ```

7. Steve said

Klong 20170905

```prime::{:[x<2; 0; :[x=2; 1; &/x!:\2+!_x^1%2]]}

nextPrime::{[a]; a::x+1; {x; prime(a)=0}{x; a::a+1}:~a; a}

findPrimes::{[max numprimes primes sum]; max::x; numprimes::sum::0; primes::[]; {x; max>numprimes}{num::nextPrime(x); numprimes::numprimes+1; sum::sum+num; :[prime(sum!numprimes)=1; primes::primes,numprimes; primes::primes]; num}:~0; primes}
```

findPrimes@365
[5 6 7 8 12 15 16 19 20 21 24 26 30 34 37 38 40 42 44 45 46 48 49 50 55 58 59 60 62 64 65 66 67 68 70 72 73 75 76 78 86 87 88 92 102 116 120 122 124 128 130 132 135 140 143 145 150 156 158 164 165 166 168 172 173 175 176 182 183 191 196 210 214 216 218 223 234 236 241 248 250 256 259 262 265 266 272 280 285 301 306 310 311 314 315 324 328 330 336 337 344 347 348 349 352 355 358 365]

8. sealfin said

October 9th, 2017.cpp:

```#include "seal_queue.h" /* <http://GitHub.com/sealfin/C-and-C-Plus-Plus/blob/master/seal_queue.h> */
#include "printDateAndTime.h" /* <http://Gist.GitHub.com/sealfin/6d35f3a3958bd6797a0f> */
#include <stdio.h>
#include <limits.h>
#include <stdlib.h>

bool *g_isPrime = NULL;

bool f_IsPrime( const unsigned short p );

bool f_IsPrime( const unsigned short p )
{
if( p % 2 == 0 )
return p == 2;
else
return g_isPrime[ p ];
}

int main( const int argc, const char * const argv[] )
{
unsigned short limit = 0;
const unsigned short k_LimitStep = 1000;
unsigned short i, number_of_primes, k, *primes, day = 1, number_of_days_sharing_characteristic = 0;
seal_Queue<unsigned short> days_sharing_characteristic;

#ifndef __MWERKS__
printf( "\n" );
#endif
p_PrintDateAndTime();
printf( "\n" );

/* Firstly, let's determine (at least) the first 366 prime numbers; we're using a (slightly optimised) sieve of Eratosthenes, but an incremental sieve of Eratosthenes (as an example) would be more appropriate. */
do
{
const unsigned int new_limit = limit + k_LimitStep;

if( new_limit > USHRT_MAX )
{
fprintf( stderr, "Sorry, an error occurred: `limit` overflowed.\n" );
if( g_isPrime != NULL)
free( g_isPrime );
exit( EXIT_FAILURE );
}
limit = new_limit;
if( g_isPrime != NULL )
free( g_isPrime );
g_isPrime = ( bool* )malloc( sizeof( bool ) * limit );
for( i = 0; i < limit; i ++ )
g_isPrime[ i ] = true;
number_of_primes = 0;
g_isPrime[ 0 ] = false;
g_isPrime[ 1 ] = false;
for( i = 3; i < limit; i += 2 )
if( g_isPrime[ i ] )
{
number_of_primes ++;
for( k = i + i; k < limit; k += i  )
g_isPrime[ k ] = false;
}
}
while( number_of_primes < 366 /* Leap year. */ );

/* Secondly, let's create a list of (at least) the first 366 prime numbers. */
k = 0;
primes = ( unsigned short* )malloc( sizeof( unsigned short ) * number_of_primes );
primes[ k ++ ] = 2;
for( i = 3; k < number_of_primes; i += 2 )
if( g_isPrime[ i ] )
primes[ k ++ ] = i;

for( ; day <= 366 /* Leap year. */; day ++ )
{
unsigned int sum_of_primes = 0;

for( i = 0; i < day; i ++ )
{
const unsigned long long new_sum_of_primes = sum_of_primes + primes[ i ];

if( new_sum_of_primes > UINT_MAX )
{
fprintf( stderr, "Sorry, an error occurred: `sum_of_primes` overflowed.\n" );
free( g_isPrime );
free( primes );
exit( EXIT_FAILURE );
}
sum_of_primes = new_sum_of_primes;
}
if( f_IsPrime( sum_of_primes % day ))
{
number_of_days_sharing_characteristic ++;
days_sharing_characteristic.p_Push( day );
}
}
free( g_isPrime );
free( primes );

printf( "There " );
if( number_of_days_sharing_characteristic != 1 )
printf( "are" );
else
printf( "is" );
printf( " " );
if( number_of_days_sharing_characteristic != 0 )
printf( "%u", number_of_days_sharing_characteristic );
else
printf( "no" );
printf( " day%s sharing the characteristic that on the Nth day, the sum of the first N primes, modulo N, is prime", ( number_of_days_sharing_characteristic != 1 )?"s":"" );
if( number_of_days_sharing_characteristic > 0 )
{
printf( ": " );
for( i = 0; i < number_of_days_sharing_characteristic; i ++ )
{
if( i > 0 )
{
if( number_of_days_sharing_characteristic != 2 )
printf( ", " );
if( i + 1 == number_of_days_sharing_characteristic )
{
if( number_of_days_sharing_characteristic == 2 )
printf( " " );
printf( "and " );
}
}
printf( "%u", days_sharing_characteristic.f_Pop());
}
}
printf( ".\n" );

printf( "\n" );
p_PrintDateAndTime();
#ifndef __MWERKS__
printf( "\n" );
#endif
exit( EXIT_SUCCESS );
}```

Output:

There are 108 days sharing the characteristic that on the Nth day, the sum of the first N primes, modulo N, is prime: 5, 6, 7, 8, 12, 15, 16, 19, 20, 21, 24, 26, 30, 34, 37, 38, 40, 42, 44, 45, 46, 48, 49, 50, 55, 58, 59, 60, 62, 64, 65, 66, 67, 68, 70, 72, 73, 75, 76, 78, 86, 87, 88, 92, 102, 116, 120, 122, 124, 128, 130, 132, 135, 140, 143, 145, 150, 156, 158, 164, 165, 166, 168, 172, 173, 175, 176, 182, 183, 191, 196, 210, 214, 216, 218, 223, 234, 236, 241, 248, 250, 256, 259, 262, 265, 266, 272, 280, 285, 301, 306, 310, 311, 314, 315, 324, 328, 330, 336, 337, 344, 347, 348, 349, 352, 355, 358, and 365.[/sourecode]

On an Apple Power Mac G4 (AGP Graphics) (450MHz processor, 1GB memory) to run the solution took approximately one second on both Mac OS 9.2.2 (International English) (the solution compiled using Metrowerks CodeWarrior IDE 2.1 (Discover Programming Edition)) and Mac OS X 10.4.11 (the solution compiled using Xcode 2.2.1).

(I’ve just completed a gig at London South Bank University and so I’m again just trying to solve the problems posed by this ‘site whilst I try to get a job (and I’ve solved this problem in particular to test my `seal_Queue` template class I might use in the fourth version of my SDL2 joystick interrogator utility); I’m well-aware that my solutions are far from the best – but, in my defence, I don’t have any traditional qualifications in computer science :/ )

Ps. Merry Christmas, and thank you for running this ‘site!