A deck of cards has 13 cards in each of four suites?

Question

Determine if total number N of cards in the deck is prime number?

Answer 1

No, the number of cards in the deck is 52, or 4x13.

Answer 2

The deck does not have a prime number of cards. You don't even have to waste your time multiplying 13 times 4 because automatically if N is divisible by 13 or 4 it couldn't be prime.

Answer 3

It can't be prime, because the total number of cards in the deck is 13 times 4 which is 52. Simply because you multiplied 13 and 4 you know it can't be prime............prime is any number that is only divisible by itself and the number 1.
52 is divisible by 52, 1, 13, 4, 2, and 26.

Answer 4

13x4= 52

and we know that 52 isn't a prime number because it is divisible by more numbers than itself and one, as we have just showed that it is divisible by 4 and 13.

so no, total number N of cards in the deck is not a prime number.

Answer 5

right it extremely is changed GetPlayValue: int GetPlayValue(int deck[NCARDS][NPROPS], int i) { int playval, tface; /* assign face fee */ tface = deck[i][a million]; if(tface==0) playval = 11; //Ace = 11 else if(tface<10) playval = tface+a million; //Ace = a million else playval = 10; //tface = a million return playval; } right it extremely is one strategies-set to place above device("PAUSE"); : // deal out 5 enjoying cards to 2 gamers int playercount[2] = {0,0}, j, thisval; for (i=0; i<5; i++) { for (j=0; j<2; j++) { thisval = GetPlayValue(deck, i*2 + j); if (thisval == a million) thisval = 11; playercount[j] += thisval; } } if (playercount[0] > playercount[a million]) { printf("participant a million wins with score = %ld %sn", playercount[0]); for (j=0; j<10; j+=2) PrintCard(deck, j); printf("participant 2 loses with score = %ld %sn", playercount[a million]); for (j=a million; j<10; j+=2) PrintCard(deck, j); } else if (playercount[0] < playercount[a million]) { printf("participant 2 wins with score = %ld %sn", playercount[a million]); for (j=a million; j<10; j+=2) PrintCard(deck, j); printf("participant a million loses with score = %ld %sn", playercount[0]); for (j=0; j<10; j+=2) PrintCard(deck, j); } else { printf("participant a million ties with score = %ld %sn", playercount[0]); for (j=0; j<10; j+=2) PrintCard(deck, j); printf("participant 2 ties with score = %ld %sn", playercount[a million]); for (j=a million; j<10; j+=2) PrintCard(deck, j); } device("PAUSE");

Answer 6

Suits, not suites.

Answer 7

YES.