Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai) - Part 16


  • NMIMS, Mumbai (Marketing)


    All pairs are formed from divisors of 21600. How many such pairs have their HCF = 45?

    21600 = 2^5 * 3^3 * 5^2.
    We need pairs with HCF = 45 = 3^2 * 5.
    Let the numbers be 45a, 45b where a, b are co-primes.
    21600/45 = 2^5 * 3 * 5 = N.
    We just need to find co-prime pairs for this N for our answer.
    This we know directly as: [(2(5) + 1) * (2(1) + 1) * (2(1) + 1) – 1]/2
    = (11 * 3 * 3 – 1)/2
    = 49 such pairs.

    How many sets of three distinct factors can be made for N = 2^6 * 3^4 * 5^2 can be made such that factors are mutually co-prime to each other in the set?

    Direct from the concepts discussed earlier.
    It is asking for unordered distinct co-prime triplets of N.
    It is given by [(3p + 1)(3q + 1)(3r + 1) – 3 * (f – 1) – 1]/6 + [Optional part not to be added here as distinct are asked]
    [(3(6) + 1) * (3(4) + 1) * (3(2) + 1) – 3 * ( 7 * 5 * 3 – 1) – 1]/6
    [19 * 13 * 7 – 3 * 104 – 1]/6
    1416/6 = 236 sets.

    X is a set of all natural numbers with 4 factors such that sum of all factors excluding the number itself is 31. Find the sum of all such possible numbers.

    If a number has exactly four factors it can be only in two forms

    Case 1
    N = a^3 where a is a prime Factors = {1, a, a^2, a^3}
    According to question(ATQ), 1 + a + a^2 = 31
    a(a + 1) = 30
    a = 5 satisfies.
    N = 125.

    Case 2
    N = a * b where a, b are distinct primes.
    Factors = {1, a, b, ab}
    ATQ, 1 + a + b = 31
    a + b = 30 where a, b are co-primes
    a, b = 7, 23 ; 11, 19; 13, 17
    N = 161, 209, 221

    Sum = 125 + 161 + 209 + 221 = 716.

    Product of four numbers is 10! Find the least possible sum of the four numbers.

    Standard Approach We know that for a constant product, sum is minimum when all the numbers are equal or as close to each other as possible.
    10! = 3628800
    Fourth root of this number would be somewhere near 45 because 45^2 = 2025 and 2025^2 = 41xxxxx > 10!
    So we will try to bring all numbers close to 45.
    10! = 2^8 * 3^4 * 5^2 * 7.
    Clearly, we can take out 45 = 3^2 * 5 leaving 2^8 * 3^2 * 5 * 7. We need other numbers near to 45 as well.
    2^3 * 5 = 40. Taking that out leaves 2^5 * 3^2 * 7.
    Take out 16 * 3 and 6 * 7 = 48 and 42 respectively.
    Minimum sum = 40 + 42 + 45 + 48 = 175.

    Smart Approach : How to figure out that we need to find numbers near to what number for such factorials?
    For factorial n, The central pivot number is found out by the following

    n = even => n!/(n/2)!*(n/2 + 1)!
    Example: n = 10 => 10! / 5! * 6! = 10 * 9 * 8 * 7/5 * 4 * 3 * 2 = 42.
    If you simply took out 42 first and tried to see numbers close to it your process will be faster and smooth.
    40, 42, 45, and 48.
    Sum = 175

    n = odd => n! / [(n + 1)/2]!^2
    Example: n = 7 => 7!/4!^2 = 7 * 6 * 5 /4 * 3 * 2 = 7 * 5/4 = 8.xx ~ 8 7! = 2^4 * 3^2 * 5 * 7
    Pivot taken out = 8. Other numbers come out easily as 7, 9, and 10.
    Sum = 34.

    S is a set of all three digit numbers. From this set, a number X is selected at random and reverse of digits of X is Y, which is also part of the same set S. What is the probability that Y is a multiple of 4, if X has already been found out as a multiple of 4?

    According to the question, if the three digit number is abc then,
    99(a – c) = 4k = 0, 4 while a, c are even.

    Case 1
    Number is aba form with a = 2, 4, 6, 8 and 5 values of b for each of the values of a.
    4 * 5 = 20 numbers

    Case 2
    2, 6; 6, 2; 4, 8; 8, 4 : 4 cases with each having 5 values of b possible
    4 * 5 = 20 numbers
    Total = 40 numbers.

    Total number of three digit numbers that are multiple of 4 = [900/4] = 225
    Probability = 40/225 = 8/45.

    Alternate : We know that the a, c are fixed by the conditions of the question while for b, its value from 0 to 9(= 10 options) will determine whether the number is divisible by 4 or not which occurs with equal probabilities.
    Hence 10/2 = 5 valid possibilities.
    a, c = 2, 2 ; 4, 4 ; 6, 6 ; 8, 8 ; 2, 6 ; 6, 2 ; 4, 8 ; 8, 4 : 8 cases.
    So total of 8 * 5 = 40 favourable cases.

    f(x) = x^(x!) and f(x) is defined for all the natural number values of x. Find the ten’s place digit of Σf(x) from x = 1 to 20.

    Last two digits of f (1) = 1^1! = 01.
    Last two digits of f (2) = 2^2! = 04.
    Last two digits of f (3) = 3^3! = 29.
    Last two digits of f (4) = 4^4! = 4^24 => 4a = 25b + 6 = 56.
    Last two digits of f (5) = 5^5! = 25.
    Last two digits of f (6) = 6^6! = 76.
    From f (5) Onwards power is a multiple of 20 so we can simply write the last two digits depending on the number being even or odd.
    f (7) = f (9) = f (11) = … = f (19) = 01.
    f (8) = f (12) = … = f (18) = 76.
    f (10) = f (20) = 00.
    f (15) = 25.

    Adding them up, [7 * 01 + 04 + 29 + 56 + 2 * 25 + 6 * 76 + 2 * 00]
    = Last two digits will be 46 + 56
    = 02.
    Hence ten’s place digit is zero.

    If the ten’s digit of a perfect square number is 7, how many units digit are possible?

    We know that there only 22 distinct possibilities of last two digits that come from squares 1-25. Rest can be calculated on the basis of these only. So a ten’s digit of 7 appears in 24^2 = 576 only.
    Hence only possible unit digit will be 6.

    Find the first and last digits of 2^43 respectively.

    Last digit of 2^43 = Last digit of 2^3 = 8.
    2,4,8,16,32,64,128,256,512,1024 2048, 4096, 8xxx, 16xxx, 32xxx, 64xxx, 128xxx, 256xxx, 512xxx, 1024xxx
    Cycles in 10 as well.
    First digit of 2^43 = First digit of 2^3 = 8.

    A number when written to the base 16 contains only 3 zeroes and three 1s and no other digits. Find the maximum number of zeroes in the number when it is represented in base 2.

    Number contains exactly three zeroes and three ones.
    Maximum and minimum of such numbers is (111000) and (100011) = (16^5 + 16^4 + 16^3) and (16^5 + 16^1 + 16^0) which in base 2 representation can be written as (2^20 + 2^16 + 2^12) and (2^20 + 2^4 + 2^0).
    In any case, the number on base 2 will be sum of three powers of 2 with the maximum power of 2 being 20.
    Hence in base 2, the number will have exactly 21 digits with exactly three 1s and (21 – 3) = 18 zeroes

    Find the last three digits of 2003^2002^2000.

    1000 = 2^3 * 5^3 = 8 * 125
    2003^2002^2000 mod 8 = 3^2002^2000 mod 8
    E (8) = 4
    2002^2000 mod 4 = 0
    3^2002^2000 mod 8 = 1.

    2003^2002^2000 mod 125 = 3^2002^2000 mod 125
    E (125) = 100
    2002^2000 mod 100 = 2^2000 mod 100 = 76(last two digits rule)
    3^76 mod 125
    (2^7)^76 mod 125 [ Remember this out of the box thinking method?]
    2^532 mod 125
    2^32 mod 125
    6^4 mod 125 [2^8 = 256, 256 mod 125 = 6]
    46

    4a + 1 = 25b + 46
    4a = 25b + 45
    a = (6b + 11) + (b + 1)/4
    b = 3 => Remainder or last three digits are 25(3) + 46 = 121 + 100k form
    Last three digits should leave remainder = 1 with 8 and 46 with 125 by divisibility rules (as 8 = 2^3 and 125 = 5^3). 121, 221, 321, 421, 521, 621, 721, 821, 921.
    921 satisfies and is our answer.

    If the sum of two natural numbers and their LCM is 89, then how many such pairs of numbers are possible?

    Let the HCF of the two numbers be h and numbers be h * a and h * b where a, b are co-primes.

    According to the question, h * a + h * b + h * a * b = 89.
    h * (a + b + ab) = 89

    Case 1
    HCF = 1
    a + b + ab = 89
    a + b + ab + 1 = 90
    (a + 1)(b + 1) = 90 = 2 * 3^2 * 5
    Number of pairs = (Factors/2) – 1{for 1, 90 case as then a = 0} = 2 * 3 * 2/2 – 1 = 5 pairs

    Case 2
    HCF not equal to 1.
    h * (a + b + ab) = 89 = prime
    Then, HCF = 89 which is not possible as sum of numbers and LCM = 89 and we know that LCM > = HCF > = Numbers always.
    So total 5 pairs of such numbers are possible

    An odometer will not register an odd digit. It always skips an odd digit and replaces it with the following even digit. What is the true reading if this faulty non-odd odometer registers 286?

    In such questions where counting while skipping certain digits is involved we always use a mapping of a new counting system and make it analogous to some base n system.
    0 < - > 0
    1 < - > 2
    2 < - > 2
    3 < - > 4
    4 < - > 4
    5 < - > 6
    6 < - > 6
    7 < - > 8
    8 < - > 8
    9 < - > 0

    That means it is an analogous base 5 system with digits to be used as 0, 2, 4, 6 and 8 instead of the usual 0, 1, 2, 3 and 4.

    The odometer’s count not only skips digits but skips the count itself. An odometer could simply be not using some digits but this odometer doesn’t register them and replaces at that instant itself the next even. So basically it runs as 0 - > 2 (1 changed to 2) - > 4 (2 + 1 = 3 changed to 4) - > 6 -> 8 - > 20 (and not 10 as would have been in a usual digit skipping system)

    So a count of 286 is basically just 286/2 = 143.

    Moreover, this count of 143 is in base 5 counting where we move the counter every 5 numbers. 0 -> 2 -> 4 -> 6 -> 8 -> 20 -> 22 -> 24 -> 26 -> 28 -> 40 -> and so on.

    So now this 143 in base 5 must be converted to base 10 count = 1 * 25 + 4 * 5 + 3 = 48.

    This is the actual distance travelled not registered properly by the faulty odometer.


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