Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai) - Part 12


  • NMIMS, Mumbai (Marketing)


    Let n be the smallest positive number such that the number S = 8^n * 5^600 has 604 digits. Then the sum of digits of S is?

    We know that 10^n has 1 followed by n zeroes which is a total of (n + 1) digits.
    S = 8^n * 5^600 = 2^3n * 5^600.
    For n = 200. S would have been 10^600 with 601 digits but we need S to have 604 digits.
    These digits can come only from additional power of 2 as power of 5 is fixed.
    Hence we look for additional power of 2 that gives us those 3 extra digits.
    2^10 = 1024 is the first number with 4 digits that gives us those 3 extra digits.
    But power of 2 is 3n that is a multiple of 3. Hence we move onwards to 2^11 and then 2^12 = 4096 [12 = 3n and 600 already = n form]
    Thus, S = 2^612 * 5^600 = 4096000..600 zeroes
    => Sum of digits = 4 + 0 + 9 + 6 = 19

    A number N when divided by a divisor D gives a remainder of 52.The number 10N when divided by D gives a remainder of 8. How many values of D are possible?

    N = aD + 52
    10N = bD + 8
    =>10N = 10aD + 520
    10N = bd + 8
    (b - 10a) * D = 512
    D is a factor of 512 and leaves a remainder of 52 => D is a factor of 512 greater than 52.
    64, 128, 256, 512: 4 values

    Let M and N be single digit numbers. If product 2M5 * 13N is divisible by 36.How many ordered pairs of (M,N) are possible?

    2M5 is odd
    => 13N is divisible by 4
    => N = 2 or 6

    2M5 * 132
    => 2M5 is divisible by 3 => M = 2, 5, 8

    ii) 2M5 * 136
    => 2M5 is divisible by 9
    => M = 2

    How many square numbers are in the sequence 11, 111, 1111, . . .?

    A square cannot end with two odd digits. So it simply follows from it that none of these will be square numbers.
    Also, 1 and 9 are the only perfect squares having all odd digits.

    Bill and Clinton take the square of a certain decimal number and express it in base 5 and 6 respectively. Then Bush comes and he takes the two representations and assuming that the expressions are in base 10, adds the numbers. Which digits cannot be the unit digits of the sum?

    Square of number in decimal can have digits : 0,1,4,5,6,9
    In Base 5 : 0,1,4
    In Base 6 : 0,1,3,4
    Sum : 0,1,2,3,4,5,7,8 possible .
    6 and 9 not possible.

    P is the product of first 30 multiples of 30. N is the total number of factors of P. In how many ways N can be written as the product of two natural numbers such that the HCF of these two natural numbers is 19?

    P = 30^30 * 30!
    Factorize this.
    2^56 * 3^44 * 5^37 * 7^4 * 11^2 * 13^2 * 17 * 19 * 23 * 29
    No of factors = 57 * 45 * 38 * 5 * 3 * 3 * 2 * 2 * 2 * 2
    19^2 * 2^a * 3^b * 5^c
    19x * 19y = Number of factors of product
    => x * y = 2^a * 3^b * 5^c
    No of ways = 2^(n-1) = 4 where n = no of distinct primes

    Find highest power of 1001 in 1001 * 999 * 997 .... 5 * 3 * 1.

    1001 = 7 * 11 * 13.
    The highest power of 1001 in the expression will be the highest power of 13 in the expression.
    We will have 13 * 1, 13 * 3, so on up to 13 * (7 * 11) in the above expression as 1001 = 13 * (7 * 11) itself.
    => All odd multiples of 13 up to 13 * (7 * 11)
    => Total multiples = (7 * 11 + 1)/2 = 39 [ Because if 1, 1 * 3, 1 * 5 there => (5 + 1)/2 = 3 multiples there]

    Also, we will have additional powers from multiples of 13^2 (=169)

    13 * (13 * 1), 13 * (13 * 3), 13 * (13 * 5) will be the multiples of 169 in above expression as 1001 = 13*(77)
    so we can have only up till 13 * (65).
    This gives us three additional powers of 13 not counted yet.
    So a total of 39 + 3 = 42 power of 13 in above expression.
    Hence highest power of 1001 in above expression = 42.

    Alternate method Multiplying and dividing by the missing numbers, i.e., 2 * 4 * 6 * 8 * ... * 1000, we will get:
    1001! / (2 * 4 * 6 * 8 * ... * 1000)
    Take 2 common from each term in denominator and it will become 1001! / (500! * 2^500)

    => Exponent of 13 will be = [Exponent of 13 in 1001! - Exponent of 13 in 500!]
    => 77 + 5 – (38 + 2)
    => 42.

    P is a prime number greater than 30. When P is divided by 30, the remainder is x. How many different values of x are possible?

    P = 30k + x
    => x = P – 30k
    Since P is a prime number greater than 30, it will be co-prime to 30.
    => P – 30k will be co-prime to 30 too.
    => x will be co-prime to 30 and less than 30 since 30 is the divisor.
    Number of possible values of x = 30(1 – 1/2)(1 - 1/3)(1 – 1/5) = 8.

    A and B play a game of ‘Dingo’ in which each player in his turn has to choose a positive integer that is less than the previous number but at least half the previous number. The player who chooses 1, loses and the game ends there. A starts by choosing 2021 and after that both the players (A & B) continue to play with the best strategy. Which integer should B choose immediately after A has chosen 2021, to ensure his own victory?

    Start from 1 and find all the winning positions, and then try to find a pattern.

    We can see that 2 is winning position for B.
    Next winning position is 5 (as A can choose 3 or 4 and in both cases B can choose 2 and then win)
    Next is 11 (as A can choose 6 or 7 or 8 or 9 or 10 and in all of these cases B can choose 5 and hence win)
    So, we can see if k is a winning position then 2k + 1 is also
    This way winning positions are 2, 5, 11, 23, 47, 95, 191, 383, 767, 1535, ....
    Answer: 1535.

    Let T be the set of integers {2, 12, 22, 32 ..., 542, 552} and S be a subset of T such that the sum of no two elements of S is divisible by 3. The maximum possible number of elements in S is?

    From 2 to 552 there are 56 numbers.
    Numbers of the form
    3n are 12, 42, 72, … 552, i.e. 19 numbers
    3n + 1 are 22, 52, 82, … 532, i.e. 18 numbers.
    3n + 2 are 2, 32, … 542, i.e. 19 numbers.
    In order that sum of any 2 numbers is not divisible by 3, Choose 19 numbers of the form 3n + 2 and 1 number of the form 3n.
    Hence, there are 20 numbers possible.


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