Quant Boosters - Hemant Malhotra - Set 9



  • 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



  • Minimum occurs when the numbers are 6, 6, 6, 2, 2, 1.
    Maximum occurs when the numbers are 7, 5, 4, 3, 2, 2.
    To achieve min LCM look for max number of hcf among the 6C2 pairs.
    So we need number in form of coprime.

    Thus, M = 414 = 2 * 3^2 * 23, M has 2 * 3 * 2 = 12 divisors
    M^2 has 3 * 5 * 3 = 45 divisors.

    so 45 - 12 = 33 divisors



  • Q13) N = 98765432109876543210.... ( 1000 digits ).What is the least positive value of n such that N + n is divisible by 11 ?



  • Divisiblity rule of 11: Numerals whose alternating sum of digits is divisible by 11 represent numbers divisible by 11. Here the "alternating sum" means we alternate the signs from positive to negative to positive to negative, and so on. 10 to an odd power plus 1 is divisible by 11, and 10 to an even power minus 1 is divisible by 11. This test can also be applied recursively

    Example: 12342
    1 - 2 + 3 - 4 + 2 = 0 so div by 11
    Another one, 2131415
    2 - 1 + 3 - 1 + 4 - 1 + 5 = 11 so div by 11

    In given case, make a group of 10 numbers and add all such group, there will be 100 groups so total sum is 9876543210 * 100 = 987654321000
    Now use alternate sum approach so remainder is -5 ; so u need to add 5



  • Q14) Exactly one of the five numbers listed below is a prime number. Which one is the prime number ?
    (a) 999,991
    (b) 999,973
    (c) 999,983
    (d) 1,000,001
    (e) 7,999,973



  • a) (10^3)^2 - 3^2 = > (1003)(997)
    b) (10^2)^3 - 3^3 = > divisible by (10^2 -3) = 97
    d) (10^2)^3 + 1^3 = > divisible by 101
    e) (2*10^2)^3 - 3^3 = > divisible by 200 -3 = 197

    so c is prime



  • Q15) Find the smallest prime number N such that the following is true:
    The largest prime factor of N − 1 is A
    The largest prime factor of A − 1 is B
    The largest prime factor of B − 1 is 7

    a) 187
    b) 347
    c) 119
    d) 311



  • use options.
    for N = 347
    346 = 2 * 173
    172 = 4 * 43
    42= 2 * 3 * 7



  • Q16) What are the number of co-primes of y less than y, where y is the largest number with which when 486, 686 and x are divided the remainders are the same and x is the largest 3 digit number which when divided by 3 or 8 leaves a remainder of 2 in each case.
    a) 10
    b) 20
    c) 30
    d) 40
    e) None of these



  • x = 24k + 2 = 986
    486, 686 and 986 gives a remainder of 86 when divided by 100
    y = 100
    co-primes = 100 * (1/2)(4/5) = 40



  • Q17) Let x and y be positive integers such that (x − y) and (x + y) are prime. Then which of the following statements is/are true?

    I. x^2 + y^2 cannot be prime.
    II. x^2 + y^2 cannot be even.
    III. x^2 − y^2 may be even or odd.

    a) I only
    b) II only
    c) III only
    d) I and III only
    e) I and II only



  • If x and y are integers, then (x − y) and (x + y) are either both even, or both odd.
    Since (x − y) and (x + y) are prime numbers, therefore, both are odd, and so one of x and y needs to be an even number and the other has to be an odd number.
    (x^2 + y^2) and (x^2 − y^2) will always be odd.
    II is true and III is false.
    Now consider x = 5 and y = 2
    x^2 + y^2 = 25 + 4 = 29 which is a prime number.
    I is false.
    so option 2



  • Q18) Find x if x^2 = 444447555561
    a) 666661
    b) 666669
    c) 233331
    d) 2111119



  • digital sum of x^2 = 9
    So digital sum of x shud be 3
    Option b's digital sum is 3 hence b.



  • Q19) Let x, y, z be prime numbers in the arithmetic progression such that x > y > z. Which among the following is always true ?
    (a) x – z is divisible by 12
    (b) x + y is divisible by 8
    (c) xz - 1 is divisible by 7
    (d) atleast two of the above
    (e) none of the above



  • x = 7, y = 5 and z = 3
    Option E



  • Q20) How many ordered and unordered triplets of non-negative integers (a , b , c) are there such that root(a) + root(b) + root(c) = root(625)


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