High School

1. A boy with a collection of marbles realizes that if he makes a group of 5 or 6 marbles, there are always two marbles left. Then p will be odd, even, prime or not prime?

2. Find the least positive integer which on adding 1 is exactly divisible by 126 and 600.

3. Find the largest possible positive integer that will divide 398, 436 and 542 leaving remainder 7, 11, 15 respectively.

Or

If A, B and C are three rational numbers such that 85C - 340A = 109, 425A + 85B = 146, then the sum of A, B and C is divisible by?

Answer :

Let's address each part of the question one by one:


  1. A boy with a collection of marbles realizes that if he makes a group of 5 or 6 marbles, there are always two marbles left. Then [tex]p[/tex] will be odd, even, prime or not prime?

    If there are always two marbles left when grouped in 5 or 6, the number of marbles [tex]p[/tex] can be expressed as:


    • [tex]p \equiv 2 \pmod{5}[/tex]

    • [tex]p \equiv 2 \pmod{6}[/tex]


    By solving these congruences, we can determine that [tex]p = 5k + 2[/tex] for some integer [tex]k[/tex]. Since [tex]p[/tex] leaves a remainder of 2 when divided by 5 and 6, it is neither strictly odd nor even, and can possibly be prime or not, depending on [tex]k[/tex]. No conclusive statement about prime-ness or parity without additional information.


  2. Find the least positive integer which on adding 1 is exactly divisible by 126 and 600.

    To find this number, let [tex]n + 1[/tex] be the number divisible by both 126 and 600. This requires calculating the Least Common Multiple (LCM) of 126 and 600.


    • First, find the prime factors:


      • [tex]126 = 2 \times 3^2 \times 7[/tex]

      • [tex]600 = 2^3 \times 3 \times 5^2[/tex]



    • The LCM will be the highest powers of all primes found:


      • [tex]\text{LCM}(126, 600) = 2^3 \times 3^2 \times 5^2 \times 7 = 12600[/tex]



    • Thus, [tex]n + 1 = 12600[/tex] and [tex]n = 12600 - 1 = 12599[/tex].




  3. Find the largest possible positive integer that will divide 398, 436 and 542 leaving remainder 7, 11, 15 respectively.

    We are tasked with finding a common divisor for these conditions:


    • [tex]d | (398 - 7), (436 - 11), (542 - 15)[/tex]

    • Simplifying gives: [tex]d | 391, 425, 527[/tex]


    Find the greatest common divisor (GCD) of these numbers:


    • [tex]391 = 17 \times 23[/tex]


    • [tex]425 = 5^2 \times 17[/tex]


    • [tex]527 = 17 \times 31[/tex]


    • Therefore, the GCD here is 17, so the largest integer is 17.





Finally, for the statement "If A, B and C are three rational numbers such that 85C - 340A = 109, 425A + 85B = 146, then the sum of A, B and C is divisible by?"


  1. We'll solve the system of equations:


    • [tex]85C - 4 \times 85A = 109 \implies 85(C - 4A) = 109[/tex]

    • [tex]5 \times 85A + 85B = 146 \implies 85(5A + B) = 146[/tex]


    Solving these, we find values for A, B, and C. Then we check the divisibility of their sum (A + B + C) by specific integers through trial and check.


Other Questions