Quant Boosters - Hemant Malhotra - Set 9



  • div by 7 ,11,13 so div by 1001
    1001=10^3+1
    so make group of 3
    and add with alternate sign from right side
    000 - abc + 674 - 307 + 1 = -abc+368 so abc=368



  • Q3) 34! = 295232799cd96041408476186096435ab000000, determine the digits a, b, c & d.



  • 34! has 7 trailing zeroes so b = 0
    first non zero digits from right side
    34! = 5 * 6+4
    so 2^6 * 6! * 4!
    4 * 2 * 4 = 2
    so a = 2
    now number is div by 7,11,13 so div by 1001 so make group 3 and add with alternate sign from right side
    000-000+5ab-643+609-618+847-140+604-cd9+799-232+295=2041-cd9
    so 2041-cd9 mod 1001=0
    so 39-cd9 mod 1001=0
    so c=0 and d=3



  • Q4) Find the number of ordered pair of digits (A,B ) such that A3640548981270644B is divisible by 99.



  • 4b + 64 + 12 + 98 + 48 + 05 + 64 + a3 =
    291 + 70 + 40 + b + 10a + 3
    404 + ab/99
    8 + ab/99
    where ab is a 2 digit number
    clearly ab can only be 91 , so only 1 pair



  • Q5) Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4



  • Any number in base k is divisible by k+1 or its divisors if and only if the sum of the digits(alternating and sign changed) is divisible by k+1 or its divisors respectively.
    like in base 10, if a number is div. by 11 then sum of its digits(sign changed alternatively) is div. by 11 and vice-versa.
    In base 7 k = 7, k+1 = 8. 4 is the divisor of 8. The alternating sum of the digits of 123abc231bca312cab is 1-2+3-a+b-c+2-3+1-b+c-a+3-1+2-c+a-b = 6 -(a +b +c) is divisible by 4
    Now (a+b+c) can be 2,6,10,14, or 18
    Solutions in each case is 6,28,36,15,1
    total solution's = 6+28+36+15+1 = 86



  • Q6) Let N = 44444444......4444444 (2045 times). find remainder when N is divided by 103 ?



  • 44444 mod 103 = 51
    any number written p - 1 times div by p where p is prime
    so only last 5 digits will be dere so 44444 mod 103=51



  • Q7) What is remainder when 2014^2015 is divided by 121 ?



  • Use Eulers theorm, 121 = 11 * 11 = > E (121)=121 * 10/11 = 110
    now 2014^2015 mod 121 = (-43)^35 mod121
    = -43 * (43 * 43)^17 mod 121
    = -43 * (34)^17 mod 121
    = -43 * 34 * (34 * 34)^8 mod 121
    = -43 * 34 * (-54 * -54)^4
    = -43 * 34 * (12)^4 mod 121
    = -43 * 34 * 144 * 23 mod 121
    = -43 * 34 * 23^2 mod 121
    = -43 * 34 * 45 Mod 121
    = -87 mod 121 = 34.

    OR

    78^2015 mod 121
    = (1+77)^2015 mod 121
    = 77C0 + 2015C1 * 77 + rest of the terms divisible by 11
    = (1 + 2015 * 77 ) mod 121
    = 34



  • Q8) If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N ?



  • highest power of 2 in N! = (2^m)-1 = N -1
    N! = 2^(N-1)*k
    N! mod 2^N = 2^(N-1) { k mod 2} =2^(N-1)



  • Q9) 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 ?
    a. 9
    b. 8
    c. 10
    d. 11



  • P = 30n + x, P has to be 6k+1 or 6k-1. Hence x also has to be of the form 6k+1 or 6k-1. Possible values of x are 1,5,7,11,13,17,19,23,25,29. out of these x cannot be 5 or 25. hence 8 values

    OR

    Just find number of coprime less than 30 = 30 * 1/2 * 2/3 * 4/5 = 8



  • Q10) If n = 1 + x, where x is the product of 4 consecutive natural numbers, then n is:
    I. an odd number.
    II. an even number.
    III. a prime number.
    IV. a perfect square.

    a) II only
    b) III and IV only
    c) I and IV only
    d) I only
    e) None of these



  • For 4 consecutive natural number there will be atleast 2 even numbers so product will be even
    = > x is even and x+1 is odd
    so 1 is true and 2nd is false
    now to decide where n is prime or not
    try for first four natural numbers which give a product of 24 and therefore n =25, therefore n is not prime but could be prefect square

    OR

    n= (a-1) * (a) * (a+1)(a+2)
    n= (a^2+a-1)(a^2+a-1)
    so n is perfect square

    OR

    x = 2 * 3 * 4 * 5.
    then 120 + 1 = 121 = 11^2

    so option 3.



  • Q11) If 2n + 1 and 3n + 1 ( n # 0 ) are perfect squares, find remainder when n is divide by 40 ?



  • n = 40 so 2n+1 = 81 = 9^2
    3n+1 = 121 = 11^2
    40 mod 40 = 0



  • Q12) The sum of 6 natural numbers (not necessarily distinct) is 23. Let M denote the positive difference of the maximum and minimum of the LCM of these 6 numbers. Then total number of divisors that divide M^2 but doesn't divide M are
    (a) 22
    (b) 33
    (c) 18
    (d) 26
    (e) none


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