Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai) - Part 13


  • NMIMS, Mumbai (Marketing)


    For how many positive integers x between 1 and 1000, both inclusive, is 4x^6 + x^3 + 5 divisible by 7?

    x^3 mod 7 = 0, 1, -1 only.
    Case 1: (4x^6 + x^3 + 5) mod 7 = 5.
    Case 2: (4x^6 + x^3 + 5) mod 7 = 4(1) + 1 + 5 mod 7 = 10 mod 7 = 3.
    Case 3: (4x^6 + x^3 + 5) mod 7 = 4(1) - 1 + 5 mod 7 = 8 mod 7 = 1.
    Hence zero values.

    n! has x no of zeroes at the end and (n+1)! has x+2 zeroes at the end. Find the no of possible values of n if n is a three digit no.

    If n! has x zeroes and (n + 1)! has (x + 2) zeroes, then n has to be of form 25k - 1 but not of form 125k - 1 (else there will be 3 zeroes)
    Number of three digit multiples of 25 = {100, 125, … , 975} = (975 – 100)/25 + 1 = 36.
    Three digit multiples of 125 = {125, 250, …, 875} = (875 – 125)/125 + 1 = 8.
    Number of possible values of n = 36 – 8 = 28.

    Product of two natural numbers = 144. What is the largest possible and smallest possible HCF of these two natural numbers.

    Largest possible HCF occurs when LCM = HCF. When LCM = HCF, numbers are equal.
    Also, product of two numbers = LCM * HCF. Since numbers have to be equal, each of the numbers = 12.
    Largest possible HCF = 12.
    Smallest possible HCF has to be obviously equal to 1, possible set for which could 144, 1.

    Product of two natural numbers = 144. How many different values of LCM are possible for these two natural numbers?

    We already calculate the largest possible HCF = 12. Subsequently smallest possible LCM = 12.
    Therefore, we can the following set of values of HCF for which corresponding LCM is also shown.
    HCF, LCM = 12, 12 ; 6, 24 ; 4, 36 ; 3, 48 ; 2, 72 ; 1, 144.
    So 6 possible LCMs.

    LCM of two natural numbers A and B = 590 and their HCF = 59. How many set of values of A and B are possible?

    Since the HCF of A and B is 59, let the numbers be A = 59x and B = 59y where x, y are co-primes.
    LCM * HCF = Product of two numbers 590 * 59 = 59x * 59y
    x * y = 10 where x, y must be co-primes.
    x, y = 1, 10; 10, 1; 2, 5; 5, 2.
    Corresponding sets of A, B = (59, 590) ; (590, 59) ; (118, 295) ; (295, 118).
    So, 4 such sets possible.

    Find the number of pairs of two natural numbers having product = 3600 and HCF = 30.

    Let the numbers be 30x, 30y where x, y are co-primes.
    30x * 30y = 3600
    x * y = 4
    x, y = (1, 4); (4, 1).
    We cannot take (2, 2) else the HCF becomes 60.
    Here since it mentions pairs only we need just the unordered sets.
    Hence, our answer = 1 set only [(1, 4)].
    NOTE: Can also be seen as number of ways to write N as a product of two co-primes = 2^(n – 1) where n is the number of distinct co-primes = 2^(1 – 1) = 1 as 4 = 2^2.

    LCM of first N natural numbers = LCM of first 10 natural numbers * 11 * 13. What can be the maximum value of N.

    N can be anything from 13 to 15 because till here neither any new primes are introduced nor an additional power of an already appearing prime is present. As soon as it becomes 16, an additional power of 2 will be required to be multiplied on RHS as LCM of first 10 natural numbers includes only power 3 of 2 through number 8.
    Hence maximum value of N = 15.

    P is a natural number such that 3 < P < 500. How many P’s are there such that HCF (P, 100) = 1.

    100 = 2^2 * 5^2.
    So, for HCF (P, 100) = 1, P must be co-prime to 2 and 5 as any power of 2 and/or 5 in P would result in a HCF of greater than 1.
    So, P belongs to (3, 500) and is co-prime to 2 and 5.
    Number of co-primes to 2 and 5 in the range 1-500 = 500 * (1 - 1/2) * (1 - 1/5) = 200[By Euler Totient function].
    Out of which P cannot be 1 or 3 as it should be greater than 3.
    2 and 500 need not be accounted for as 200 already doesn’t include them.
    Hence we have 200 – 2 = 198 valid values of P.

    LCM of 12^24, 16^18 and N is 24^24. Number of all possible values of N = S. What is the value of S?

    12^24 = (2^2 * 3)^24 = 2^48 * 3^24
    16^18 = (2^4)^18 = 2^72
    LCM = 24^24 = (2^3 * 3)^24 = 2^72 * 3^24
    LCM takes into accounts highest powers of all primes that appear in any of the participating numbers.
    Since highest power of LCM required for each prime already exists in the provided numbers, N can have any powers of 2 and 3 in its composition as long as they don’t exceed the respective powers of LCM.
    So, N = 2^p * 3^q where p = 0 to 72 and q = 0 to 24.
    S = Number of values of N = 73 * 25 = 73 * (100/4) = 7300/4 = 1825.

    What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)?

    The two given numbers are two consecutive even numbers hence their HCF = 2[Assume any two consecutive even numbers to see this].
    Also, LCM * HCF = Product of two numbers.
    Unit digit of (3^2003 – 1) = Unit digit of (3^3) - Unit digit of (1) = 7 – 1 = 6.
    Unit digit of (3^2003 + 1) = Unit digit of (3^3) + Unit digit of (1) = 7 + 1 = 8.
    Unit digit of LCM * Unit digit of HCF = ….6 * ….8
    Unit digit of LCM * 2 = 8
    Unit digit of LCM = 4.
    NOTE: It is wrong to say that LCM of unit digit of numbers is LCM (6, 8) = 24 which has unit digit as 4. Though, it comes to be correct in this case, it is not mathematically correct and should not be followed.


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