[Math Game] The Four Fours

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@markr,

Z[ P(P( T(4) + P(4!!) )) ] = Z[ P(P(10*19)) ] = Z[9293]

Covers 8893 to 9603 because of the new Z[...] function described on p.6

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@Kolyo,

Z[φ(P(P(P(√4))))^4] = Z[10^4] = Z[10000]

Covers 9600 to 10400 because of the new Z[...] function described on p.6

1 Reply

@markr,

= Z[P(P(P(P([P(P(√4))]!!))) - 4!)] = Z[P(P(P(P(15))) - 4!)] = Z[P(P(211) - 4!)] = Z[P(1297 - 4!)] = Z[P(1273)] = Z[10,391]

Covers 9,991 to 10,791 because of the new Z[...] function described on p.6

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If anyone wants to join the game here's some Java code to give you the first however many values of the P(-) function. (Recall from earlier posts, that P(n) = the nth prime.) At present this just gives the first 2000 values of P(n). Change the "2000" to equal the number of primes you want.

Code:



public class FigurePrimes {

	public static final int ORDER = 2000;

	public static void main(String[] args) {

		System.out.println("n\tnth prime\n---\t---\n1\t2");

		int n = 3;

		for (int i=2; i<=ORDER; i++){

			while (!isPrime(n)) n += 2;

			System.out.println(i + "\t" + n);

			n += 2;

		}

	}

	

	public static boolean isPrime(int n){

		for (int d=2; d*d <= n; d++){

			if (n%d==0) return false;

		}

		return true;

	}

}

Also, FYI, φ(-) is Euler's totient function, and T(n) is the nth triangular number.
(Two other common functions we use.)

0 Replies

@Kolyo,

Can you give the formula how you get this answer? Thank you.

1 Reply

@MattieSmith,

You want a formula for "5 = (4 * 4 + 4) / 4 " ?
(That's the content of the specific post that you replied to.)

Well, just remember PEMDAS for the order of arithmetic operations:
parentheses -> exponents -> multiplication & division -> addition & subtraction.

In the expression, first evaluate what's in the parentheses first, and do the multiplication before the addition, so we have: (4*4 + 4) = (16 + 4) = 20.
So we just have to divide that by 4 to get (4 * 4 + 4) / 4 = 20 / 4 = 5.

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[Math Game] The Four Fours

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