Quant Boosters - Anubhav Sehgal, NMIMS Mumbai - Set 1


  • NMIMS, Mumbai (Marketing)


    Number of Questions - 30
    Topic - Number Theory
    Solved ? - Yes
    Source -


  • NMIMS, Mumbai (Marketing)


    Q1) Two numbers when divided by m (m < = 7) gives remainder 3 and 4. The remainder when the sum of these two numbers is divided by the same divisor is R. Find the value of R + m.


  • NMIMS, Mumbai (Marketing)


    Let the two numbers be a and b
    a = mq + r1
    b = nq + r2
    Then, (a + b) = (m + n)q + (r1 + r2)
    Remainder when sum of two numbers (giving individual remainders r1 and r2 with a divisor) is divided by the same divisor is :
    R = (r1 + r2) - divisor [if this is positive]
    else
    R = (r1 + r2)
    Here R = (3 + 4) - m
    R + m = 7


  • NMIMS, Mumbai (Marketing)


    Q2) Calculate GCD(1,155) + GCD(2,155) + .. + GCD(155,155). where GCD(m,n) is the greatest common divisor of m and n.


  • NMIMS, Mumbai (Marketing)


    GCD(x,155) = 1 unless x is a multiple of 5 and/or 31
    4 only-multiples of 31
    30 only-multiples of 5
    1 multiple of both 31 and 5
    35 such numbers
    Hence,
    Sum = 4 * 31 + 30 * 5 + 1 * 155 + (155 - 35) * 1 = 549

    This is an example of how test setters combine two simple concepts to trap you.


  • NMIMS, Mumbai (Marketing)


    Q3) Find the largest number n such that (n - 11) divides n^3 + 83.


  • NMIMS, Mumbai (Marketing)


    A mix of elementary algebra and number system
    By remainder theorem,
    Remainder when n^3 + 83 is divided by (n - 11) = 11^3 + 83 = 1414
    Now,
    (n - 11) * k = 1414
    For largest n,
    k = 1, (n - 11) = 1414
    n = 1425


  • NMIMS, Mumbai (Marketing)


    Q4) Consider natural numbers a, b, c, d and e.Three statements are provided:
    I. (de)^2 = abbb
    II. (ee)^2 = ccbb
    III. A perfect square never ends with c,d and e.

    Which of the following statements are sufficient to determine the arithmetic mean of a, b and c?
    a) Any of the three statements
    b) Any two of the three statements
    c) Statement I and II are sufficient
    d) Statement II and III are sufficient
    e) All three statements are necessary


  • NMIMS, Mumbai (Marketing)


    Square with largest occurrence of a non-zero digit at the end is 38^2 = 1444
    The only square of the form XX^2 = YYZZ is 88^2 = 7744
    A square never ends with 2,3,7,8 or odd number of zeroes
    Combine them, analyze the options and you ll find I and II are sufficient to answer what we need which is primarily values of a,b,c.


  • NMIMS, Mumbai (Marketing)


    Q5) Find 24^1202 mod 1446.


  • NMIMS, Mumbai (Marketing)


    1446 = 2 * 3 * 241
    When we cancel out some common term from dividend and divisor, we need to multiply it back in the end.
    24^1202 = (2^3 * 3)^1202 = 2^3606 * 3^1202
    2^3605 * 3^1202 mod 3*241

    Use Chinese remainder theorem now,
    N = 2^3605 * 3^1202 N mod 3 = 0
    E(241) = 240
    N mod 241 = 2^3605 * 3^1202 mod 241 = (2^5 * (2^240)^15) * (3^2 * (3^240)^5) mod 241
    [Taking out powers in multiple of Euler number, as they would leave remainder as 1 with 241.]
    N mod 241 = 2^5 * 3^2 mod 241 = 47
    3a = 241b + 47
    a = (240b + 45)/3 + (b + 2)/3
    a = 80b + 15 + (b + 2)/3
    b = 1, a = 96
    Remainder = 288
    Final remainder = 288*2 = 576 [ Remember we needed to multiply the 2 canceled out earlier back]

    Alternate approach
    We know that for a prime p, N^(p – 1) mod p = 1 ; By extension of Euler theorem which is also called Fermat’s remainder theorem. That is to say, Cyclicity of obtaining remainder 1 is E (p).
    Extending that logic, 1446 = 2 * 3 * 241
    Cyclicity of 1446 = LCM (E (2), E (3), E (241)) = LCM (1, 2, 240) = 240
    24^1202 mod 1446 = (24^240)^5 * 24^2 mod 1446 = 1 * 576 mod 1446 = 576


  • NMIMS, Mumbai (Marketing)


    Q6) The product of 3 numbers a, b and c is 12 times their HCF. Find the number of ordered triplets (a,b,c) satisfying this condition?


  • NMIMS, Mumbai (Marketing)


    Numbers : hp,hq,hr where h is the hcf
    h^3. pqr = 12h
    h^2. pqr = 12
    h = 1 => pqr = 12 => 18 ordered sets
    h = 2 => pqr = 3 => 3 ordered sets
    Total : 21 ordered sets


  • NMIMS, Mumbai (Marketing)


    Q7) Find the highest N such that 11^N divides (97! + 98! + 99!)


  • NMIMS, Mumbai (Marketing)


    OA : 10, Guess doesn't need any discussion here :)


  • NMIMS, Mumbai (Marketing)


    Q8) When a number N(N > 500) is successively divided by 3,5,7, it leaves a remainder of 2,3,6 respectively. Find the remainder when 2nd such smallest N is divided by 35.


  • NMIMS, Mumbai (Marketing)


    N = 3(5(7m + 6) + 3) + 2 = 105m + 101
    N = 101,206,311,416,521,626
    626 mod 35 = 31


  • NMIMS, Mumbai (Marketing)


    Q9) Find the highest power of 1001 in 1001 * 999 * 997 * 995 * ... * 3 * 1


  • NMIMS, Mumbai (Marketing)


    Approach 1 :
    Multiply and divide by 2 * 4 * 6 * 8 *... * 998 * 1000
    Number = 1001!/(2 * 4 * 6 * 8 * .. * 998 * 1000) = 1001!/2^500 * 500!
    Exponent of 13 in this = (77 + 5) - (38 + 2) = 42

    Approach 2 :
    For power of 13, 13 * 1 to 13 * (7 * 11) all odd powers
    So, (7 * 11 + 1)/2 = 39
    For 13^2,
    13 * (13 * 1), 13 * (13 * 3), ... up-to 13 * (13 * 5)
    So additional three powers.
    Total = 39 + 3 = 42


  • NMIMS, Mumbai (Marketing)


    Q10) If p, q, and r are positive integers such that (p + q - r)(q + r - p)(p + r - q) = 15, then what is the product of p, q and r?
    a) 24
    b) 60
    c) 64
    d) Cannot Be Determined


 

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