Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai) - Part 9


  • NMIMS, Mumbai (Marketing)


    43^444 + 34^333 is divisible by?
    a) 2
    b) 5
    c) 9
    d) 11

    Looking at the options, finding the unit digit will tell us divisibility for two of them (2 and 5).
    Unit digit of 43^444 = Unit digit of 3^444 = Unit digit of 3^4 = 1
    Unit digit of 34^333 = Unit digit of 4^333 = Unit digit of 4^1 = 4
    Unit digit of expression = 1 + 4 = 5 => number is divisible by 5.

    The alternate way would have been to find remainder for each of the options.
    a) 43^444 + 34^333 mod 2 = 1^444 + 0 mod 2 = 1; Not divisible.
    b) 43^444 + 34^333 mod 9
    E (9) = 6
    (-2)^444 + (-2)^333 mod 9
    444 mod 6 = 0
    333 mod 6 = 3
    so we get, 1 + (-2)^3 mod 9 = -7 or 2.
    Not divisible
    c) 43^444 + 34^333 mod 11
    (-1)^444 + 1^333 mod 11
    1 + 1 mod 11 = 2.
    Not divisible.

    Find the remainder when 1 × 2 + 2 × 3 + 3 × 4 + ... + 98 × 99 + 99 × 100 is divided by 101.

    1 * 2 + 2 * 3 + 3 * 4 + … + 99 * 100 mod 101
    2 + 6 + 12 +20 + … + 9900 mod 101
    Let’s look at this series.
    2 ---- (4) ---- 6 ---- (6) ----- 12 ----- (8) ----- 20 ------ (10) ----- 30 ---
    The differences of the series are in AP. So this is called as a level 1 AP. General term for a level 1 AP is taken as a quadratic equation(cubic for level 2 and so on).
    Tn = an^2 + bn + c

    T1 = a + b + c = 2
    T2 = 4a + 2b + c = 6
    T3 = 9a + 3b + c = 12

    T2 – T1 = 3a + b = 4
    T3 – T2 = 5a + b = 6
    2a = 2
    a = 1
    3a + b = 4 => b = 1
    a + b + c = 2 => c = 0

    Hence, Tn = n^2 + n
    Sn = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)/2 * [(2n + 1)/3 + 1] = n(n + 1)(n + 2)/3

    Our expression = S99 = 99 * 100 * 101/3 = 33 * 100 * 101 (Expression) mod 101 is therefore zero.

    Find remainder when 1 * 1! + 2 * 2! + 3 * 3! + … + 10 * 10! is divided by 11.

    1 * 1! + 2 * 2! + … + 10 * 10! Mod 11
    (2 – 1)1! + (3 – 1)2! + …. + (11 – 1)10! Mod 11
    2 * 1! – 1 * 1! + 3 * 2! – 1 * 2! + … + 11 * 10! – 1 * 10! Mod 11
    11 * 10! – 1 * 1! Mod 11 [Rest all terms cancel out]
    11! – 1 mod 11 = -1 or 10.

    Find the remainder when 1^39 + 2^39 + 3^39 + 4^39 + ... + 12^39 is divided by 39.

    We know that (a^n + b^n + … + k^n) is divisible by (a + b + c + .. + k) when n is odd and a, b, c, .. , k are in AP.
    So, our expression is divisible by 12 * 13/2 = 78 and hence by extension by 39 since 39 is a factor of 78.
    Remainder = 0.

    Find remainder when 23^23 is divided by 53.

    23^2 mod 53 = -1
    23^23 mod 53 = (23^2)^11 * 23 mod 53 = -1^11 * 23 mod 53 = -23 or 30.

    Sum of four two digit numbers is 256. None of the four numbers have a digit zero. Also all the eight digits are different. Which digit has not been used in the four numbers?

    Here we use a new concept called the digital sum concept.
    Digital sum is the recursive sum of digits until a single digit is obtained.
    Example: 149 = 1 + 4 + 9 = 14 = 1 + 4 = 5. 5 is the digital sum of 149.

    Important usage of this concept

    1. To check calculations. The digital sum must be same for your addition/subtraction/multiplication on both sides of your equation.
    2. Multiplying any number of 9 makes it digital sum of the result to be always 9.
      Example: 142 * 9 = 1278 = 1 + 2 + 7 + 8 = 18 = 1 + 8 = 9.
      Also, see here that, 1 + 4 + 2 = 7. 7 * 9 = 63 = 6 + 3 = 9 = Same as our result 1278’s digital sum.

    Let us see how we will use this concept here to solve our problem.
    We know that sum of four two digit numbers is 256 = 2 + 5 + 6 = 13 = 1 + 3 = 4 [digital sum].
    Also, none of the numbers uses zero while all digits are distinct. Means we use 8 out of the 9 available digits as zero isn’t used at all.
    Had we used all 9 digits somewhere, we would have had digital sum of LHS as (1 + 2 + .. + 9) = 4 * 9 = 9.
    Due to the missing digit we are getting the digital sum as 4.
    Hence the missing digit, x is given by 9 – x = 4 => x = 5.

    What is the unit digit of [10^3000/(10^100 + 3)] where [.] is GIF.

    Find remainder of (10^3000/(10^100 + 3))
    Put x = 10^100 => x^30/(x + 3).
    Put x = -3 => (-3)^30 = 3^30 [Remainder Theorem]
    Dividend = Quotient*Divisor + Remainder
    Check for unit digit on both sides,
    0 = x * 3 + 9 [ Dividend = 10^3000, Divisor = 10^100 + 3, Remainder = 3^30 , Quotient = unknown]
    x = 7 [ Quotient’s unit digit is our answer]

    Find the 50th digit after decimal in the expansion of 1/73.

    This question is an extension to the concept we discussed in the previous question.
    Say I have, 1/17 = 0.0588.... and I need the 4th digit after decimal. Multiplying it by 10^4, we get, 588.xxxx and unit digit of the integral part of this number is our answer for this question.
    So what you now need here basically is:
    Unit digit of [10^50/73] where [.] is GIF. This is now same as the previous question for you.
    Find remainder for 10^50 when divided by 73.
    Another important point here is that you should learn prime factorizations of 10^n – 1 and 10^n + 1
    numbers upto n = 5 or 6. Like 999 = 27 * 37 and 1001 = 7 * 11 * 13.

    Like here, we have, 10001 = 10^4 + 1 with 73 as its factor.
    So 10^4 mod 73 = -1
    10^50 mod 73 = (10^4)^12 * 10^2 mod 73 = 1 * 100 mod 73 = 27.
    Unit digit of remainder = 7.
    Dividend = Divisor*Quotient + Remainder
    Equate unit digit on both sides
    0 = Q * 3 + 7
    Q = 1 = Your answer.

    Find the number of integers, N such that (N^2 + 2N - 8)/(N^2 + N - 12) is an integer.

    N^2 + 2N - 8 = (N - 2)(N + 4)
    N^2 + N - 12 = (N - 3)(N + 4)
    => (N - 2)/(N - 3) is an integer
    => (N - 3 + 1)/(N - 3) is an integer
    => 1 + [1/(N - 3)] is an integer
    => N = 4. Only one possible value.

    Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2,5 or 6.

    Standard Approach
    _ : 6 numbers ;
    _ _ : 6 * 7 = 42 numbers ;
    _ _ _ : 6 * 7 * 7 = 294 numbers
    342 numbers added till 999
    1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now.
    Remaining 1000 - 685 = 315
    3 0 _ _ : 49 added
    3 1 _ _ : 49 added
    3 3 _ _ : 49 added
    3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49 * 3 = 147
    Total 294 more. Remaining 21
    3 9 0 _ : 7
    3 9 1 _ : 7
    3 9 3 _ : 7
    3939

    Smart Approach[Using base system]
    Analogous mapping of digits of this new system where 2,5,6 are absent unlike the known base 7 where 7,8,9 are not there. Remember base system is just a count. You can always map it to different digits and count differently. Only problem people face here is because we are too used to counting from 1 to 9 and further.
    Mapping the available digits to the normal count, we have
    0 -> 0
    1 -> 1
    2 -> 3
    3 -> 4
    4 -> 7
    5 -> 8
    6 -> 9
    1000th term of our series will thus be 1000 [Read as one zero zero zero] in base 7 = 2626
    Mapping it back to original digits < - > 3939.
    [1000 = 2 * 343 + 6 * 49 + 2 * 7 + 6 * 1. Hence 1000 = 2626 in base 7.]


Log in to reply
 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.