Quant Boosters - Anubhav Sehgal, NMIMS Mumbai - Set 5



  • 36
    1 * 1 * 36 : Sum = 38
    1 * 2 * 18 : Sum = 21
    1 * 3 * 12 : Sum = 16
    1 * 4 * 9 : Sum = 14
    1 * 6 * 6 : Sum = 13
    2 * 2 * 9 : Sum = 13
    2 * 3 * 6 : Sum = 11
    3 * 3 * 4 : Sum 10

    When he knows the product of ages he can come up with all these combinations. After having a look at house number, he still needs more information. This means that house number which is the sum of ages is not unique for the combinations he has. That leaves us with: 1 * 6 * 6 : Sum = 13 and 2 * 2 * 9 : Sum = 13

    Now the owner says that his “youngest” child is sleeping upstairs. It means we cannot have 2, 2, 9 as our combination as there won’t be a “youngest” child then but two younger children. Hence the individual ages are 1, 6, 6 while the house number = 13.



  • Q10) Given that 34! = X95232799ab96041408476186096435cd000000 where X is the leading digit and a, b, c and d are missing digits from the actual value of 34! What is the sum of (a + b + c + d)?
    a) 0
    b) 1
    c) 4
    d) 6
    e) none of these



  • cd can be found out with the help of last two non-zero digits of 34!
    34 = 5(6) + 4.
    Last two non-zero digits = 12^6 * 6! * 31 * 32 * 33 * 34 = 84 * 72 * 24 = 84 * 3 * 76 = 52 * 76 = 52.
    Given the options you could directly mark none of these.

    Otherwise, let us see how to proceed. Choose any three primes in the range 1- 34 whose divisibility test will involve a, b and X. Form equations and solve for the three variables.
    You will get a = 0, b = 3 and X = 2.
    (a + b + c + d) = 10



  • Q11) All pairs are formed from divisors of 21600. How many such pairs have their HCF = 45?



  • 21600 = 2^5 * 3^3 * 5^2.
    We need pairs with HCF = 45 = 3^2 * 5.
    Let the numbers be 45a, 45b where a, b are co-primes.
    21600/45 = 2^5 * 3 * 5 = N.
    We just need to find co-prime pairs for this N for our answer.
    This we know directly as: [(2(5) + 1) * (2(1) + 1) * (2(1) + 1) – 1]/2
    = (11 * 3 * 3 – 1)/2
    = 49 such pairs.



  • Q12) How many sets of three distinct factors can be made for N = 2^6 * 3^4 * 5^2 can be made such that factors are mutually co-prime to each other in the set?



  • Direct from the concepts discussed earlier.
    It is asking for unordered distinct co-prime triplets of N.
    It is given by [(3p + 1)(3q + 1)(3r + 1) – 3 * (f – 1) – 1]/6 + [Optional part not to be added here as distinct are asked]
    [(3(6) + 1) * (3(4) + 1) * (3(2) + 1) – 3 * ( 7 * 5 * 3 – 1) – 1]/6
    [19 * 13 * 7 – 3 * 104 – 1]/6
    1416/6 = 236 sets.



  • Q13) X is a set of all natural numbers with 4 factors such that sum of all factors excluding the number itself is 31. Find the sum of all such possible numbers.



  • If a number has exactly four factors it can be only in two forms

    Case 1
    N = a^3 where a is a prime Factors = {1, a, a^2, a^3}
    According to question(ATQ), 1 + a + a^2 = 31
    a(a + 1) = 30
    a = 5 satisfies.
    N = 125.

    Case 2
    N = a * b where a, b are distinct primes.
    Factors = {1, a, b, ab}
    ATQ, 1 + a + b = 31
    a + b = 30 where a, b are co-primes
    a, b = 7, 23 ; 11, 19; 13, 17
    N = 161, 209, 221

    Sum = 125 + 161 + 209 + 221 = 716



  • Q14) Product of four numbers is 10! Find the least possible sum of the four numbers.



  • Standard Approach We know that for a constant product, sum is minimum when all the numbers are equal or as close to each other as possible.
    10! = 3628800
    Fourth root of this number would be somewhere near 45 because 45^2 = 2025 and 2025^2 = 41xxxxx > 10!
    So we will try to bring all numbers close to 45.
    10! = 2^8 * 3^4 * 5^2 * 7.
    Clearly, we can take out 45 = 3^2 * 5 leaving 2^8 * 3^2 * 5 * 7. We need other numbers near to 45 as well.
    2^3 * 5 = 40. Taking that out leaves 2^5 * 3^2 * 7.
    Take out 16 * 3 and 6 * 7 = 48 and 42 respectively.
    Minimum sum = 40 + 42 + 45 + 48 = 175.

    Smart Approach : How to figure out that we need to find numbers near to what number for such factorials?
    For factorial n, The central pivot number is found out by the following

    n = even => n!/(n/2)!*(n/2 + 1)!
    Example: n = 10 => 10! / 5! * 6! = 10 * 9 * 8 * 7/5 * 4 * 3 * 2 = 42.
    If you simply took out 42 first and tried to see numbers close to it your process will be faster and smooth.
    40, 42, 45, and 48.
    Sum = 175

    n = odd => n! / [(n + 1)/2]!^2
    Example: n = 7 => 7!/4!^2 = 7 * 6 * 5 /4 * 3 * 2 = 7 * 5/4 = 8.xx ~ 8 7! = 2^4 * 3^2 * 5 * 7
    Pivot taken out = 8. Other numbers come out easily as 7, 9, and 10.
    Sum = 34.



  • Q15) S is a set of all three digit numbers. From this set, a number X is selected at random and reverse of digits of X is Y, which is also part of the same set S. What is the probability that Y is a multiple of 4, if X has already been found out as a multiple of 4?



  • According to the question, if the three digit number is abc then,
    99(a – c) = 4k = 0, 4 while a, c are even.

    Case 1
    Number is aba form with a = 2, 4, 6, 8 and 5 values of b for each of the values of a.
    4 * 5 = 20 numbers

    Case 2
    2, 6; 6, 2; 4, 8; 8, 4 : 4 cases with each having 5 values of b possible
    4 * 5 = 20 numbers
    Total = 40 numbers.

    Total number of three digit numbers that are multiple of 4 = [900/4] = 225
    Probability = 40/225 = 8/45.

    Alternate : We know that the a, c are fixed by the conditions of the question while for b, its value from 0 to 9(= 10 options) will determine whether the number is divisible by 4 or not which occurs with equal probabilities.
    Hence 10/2 = 5 valid possibilities.
    a, c = 2, 2 ; 4, 4 ; 6, 6 ; 8, 8 ; 2, 6 ; 6, 2 ; 4, 8 ; 8, 4 : 8 cases.
    So total of 8 * 5 = 40 favourable cases.



  • Q16) f(x) = x^(x!) and f(x) is defined for all the natural number values of x. Find the ten’s place digit of Σf(x) from x = 1 to 20.



  • Last two digits of f (1) = 1^1! = 01.
    Last two digits of f (2) = 2^2! = 04.
    Last two digits of f (3) = 3^3! = 29.
    Last two digits of f (4) = 4^4! = 4^24 => 4a = 25b + 6 = 56.
    Last two digits of f (5) = 5^5! = 25.
    Last two digits of f (6) = 6^6! = 76.
    From f (5) Onwards power is a multiple of 20 so we can simply write the last two digits depending on the number being even or odd.
    f (7) = f (9) = f (11) = … = f (19) = 01.
    f (8) = f (12) = … = f (18) = 76.
    f (10) = f (20) = 00.
    f (15) = 25.

    Adding them up, [7 * 01 + 04 + 29 + 56 + 2 * 25 + 6 * 76 + 2 * 00]
    = Last two digits will be 46 + 56
    = 02.
    Hence ten’s place digit is zero.



  • Q17) If the ten’s digit of a perfect square number is 7, how many units digit are possible?



  • We know that there only 22 distinct possibilities of last two digits that come from squares 1-25. Rest can be calculated on the basis of these only. So a ten’s digit of 7 appears in 24^2 = 576 only.
    Hence only possible unit digit will be 6.



  • Q18) Find the first and last digits of 2^43 respectively.



  • Last digit of 2^43 = Last digit of 2^3 = 8.
    2,4,8,16,32,64,128,256,512,1024 2048, 4096, 8xxx, 16xxx, 32xxx, 64xxx, 128xxx, 256xxx, 512xxx, 1024xxx
    Cycles in 10 as well.
    First digit of 2^43 = First digit of 2^3 = 8.



  • Q19) A number when written to the base 16 contains only 3 zeroes and three 1s and no other digits. Find the maximum number of zeroes in the number when it is represented in base 2.


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