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Question: two numbers p and q are randomly selected from a set of natural numbers. then probability that a^2+b...

two numbers p and q are randomly selected from a set of natural numbers. then probability that a^2+b^2 is divisible by 5 is

A

1/5

B

9/25

C

4/25

D

1/25

Answer

9/25

Explanation

Solution

The problem requires finding the probability that a2+b2≡0(mod5)a^2 + b^2 \equiv 0 \pmod{5} for randomly selected natural numbers aa and bb.

  1. Analyze Remainders of Squares Modulo 5: The possible remainders when a natural number is divided by 5 are 0, 1, 2, 3, and 4. The squares of these remainders modulo 5 are:

    • 02≡0(mod5)0^2 \equiv 0 \pmod{5}
    • 12≡1(mod5)1^2 \equiv 1 \pmod{5}
    • 22≡4(mod5)2^2 \equiv 4 \pmod{5}
    • 32≡9≡4(mod5)3^2 \equiv 9 \equiv 4 \pmod{5}
    • 42≡16≡1(mod5)4^2 \equiv 16 \equiv 1 \pmod{5} So, the possible remainders for n2(mod5)n^2 \pmod{5} are {0, 1, 4}.

  2. Determine Probabilities of Square Remainders: Assuming natural numbers are uniformly distributed across remainders modulo 5, the probability of a number having any specific remainder is 1/51/5.

    • P(n2≡0(mod5))=P(n≡0(mod5))=1/5P(n^2 \equiv 0 \pmod{5}) = P(n \equiv 0 \pmod{5}) = 1/5.
    • P(n2≡1(mod5))=P(n≡1(mod5) or n≡4(mod5))=1/5+1/5=2/5P(n^2 \equiv 1 \pmod{5}) = P(n \equiv 1 \pmod{5} \text{ or } n \equiv 4 \pmod{5}) = 1/5 + 1/5 = 2/5.
    • P(n2≡4(mod5))=P(n≡2(mod5) or n≡3(mod5))=1/5+1/5=2/5P(n^2 \equiv 4 \pmod{5}) = P(n \equiv 2 \pmod{5} \text{ or } n \equiv 3 \pmod{5}) = 1/5 + 1/5 = 2/5. These probabilities apply independently to both aa and bb.

  3. Identify Favorable Combinations: We need a2+b2≡0(mod5)a^2 + b^2 \equiv 0 \pmod{5}. Let Ra=a2(mod5)R_a = a^2 \pmod{5} and Rb=b2(mod5)R_b = b^2 \pmod{5}. The pairs (Ra,Rb)(R_a, R_b) that sum to 0(mod5)0 \pmod{5} are:

    • (0,0)(0, 0): 0+0=00 + 0 = 0
    • (1,4)(1, 4): 1+4=5≡0(mod5)1 + 4 = 5 \equiv 0 \pmod{5}
    • (4,1)(4, 1): 4+1=5≡0(mod5)4 + 1 = 5 \equiv 0 \pmod{5}

  4. Calculate Total Probability: The probability of each case is calculated using independence:

    • P(Ra=0 and Rb=0)=P(Ra=0)×P(Rb=0)=(1/5)×(1/5)=1/25P(R_a=0 \text{ and } R_b=0) = P(R_a=0) \times P(R_b=0) = (1/5) \times (1/5) = 1/25.
    • P(Ra=1 and Rb=4)=P(Ra=1)×P(Rb=4)=(2/5)×(2/5)=4/25P(R_a=1 \text{ and } R_b=4) = P(R_a=1) \times P(R_b=4) = (2/5) \times (2/5) = 4/25.
    • P(Ra=4 and Rb=1)=P(Ra=4)×P(Rb=1)=(2/5)×(2/5)=4/25P(R_a=4 \text{ and } R_b=1) = P(R_a=4) \times P(R_b=1) = (2/5) \times (2/5) = 4/25. The total probability is the sum of these mutually exclusive probabilities: P(a2+b2≡0(mod5))=1/25+4/25+4/25=9/25P(a^2 + b^2 \equiv 0 \pmod{5}) = 1/25 + 4/25 + 4/25 = 9/25.