Quant Boosters - HandaKaFunda - Set 3


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    What is the units digit of 2! +4! +6! ...+98!?

    We need to find out the units digits of
    2! + 4! + 6! + .... 98!
    Let us look at the units digit for all of them
    2! = 1 * 2 = 2; unit's digit is 2
    4! = 1 * 2 * 3 * 4 = 24; unit's digit is 4
    6! = 1 * 2 * 3 * 4 * 5 * 6 = 720; unit's digit is 0
    8! = multiple of 6!, divisible by 10; unit's digit is 0
    10! = multiple of 6!, divisible by 10; unit's digit is 0
    12! = multiple of 6!, divisible by 10; unit's digit is 0
    .
    .
    .
    98! = multiple of 6!, divisible by 10; unit's digit is 0

    Unit's digit of 2! + 4! + 6! + .... 98!
    = Sum of unit's digits of individual terms
    = 2 + 4 + 0 + 0 + 0 ..... 0
    = 6

    What is the remainder when 97! is divided by 101?

    Wilson's Theorem says For a prime number 'p'
    Rem [ (p-1)! / p] = p-1

    This can be extended to say,
    Rem [ (p-2)! / p] = 1

    Let us use that here. We need to find out Rem [97! / 101] = r
    We know from the above theorem,
    Rem [99! / 101] = 1
    => Rem [99 * 98 * 97! / 101] = 1
    => Rem [ (-2)*(-3)*r / 101] = 1
    => Rem [6r / 101] = 1
    => 6r = 101k + 1
    We need to think of a value of k, such that 101k + 1 is divisible by 6.
    If we put, k = 1, we get 101 + 1 = 102, which is divisible by 6.
    => 6r = 102
    => r = 17

    If a person starts writing all 4 digit numbers, how many times has he written the digit 2?

    Let us call the 4 digit number as abcd
    When a = 2, b / c / d can be filled in 10 ways each.
    So, 2 will appear in place of 'a' 10 * 10 * 10 = 1000 times
    When b = 2, a can be filled in 9 ways whereas c / d can be filled in 10 ways each.
    So, 2 will appear in place of 'b' 9 * 10 * 10 = 900 times
    Similarly, 2 will appear in place of 'c' 900 times and in place of 'd' 900 times.
    Total number of times digit '2' will appear = 1000 + 900 + 900 + 900 = 3700 times

    What is the number of digits in 2^2009?

    Using Logarithms is probably the fastest and the best way to get to the answer.
    If Log(x) = a.bcde, number of digits in x is a + 1
    Log (2^2009) = 2009 Log(2) = 2009*0.3010 = 604.709
    => 2^2009 has 604 + 1 = 605 digits

    How many digits are there in (2ABC)^4 where 2ABC is a 4 digit number?

    We need to find out the number of digits in (2ABC)^4
    The smallest such number will be 2000^4 = 16*10^12
    This will have 14 digits.

    The biggest such number will be 2999^4. While that is hard to find out, we can find out 3000^4 easily.
    3000^4 = 81*10^12. It has 14 digits
    => 2999^4 will also have 14 digits.

    As the smallest and the biggest number of the format (2ABC)^4 has 14 digits, we can say that there are 14 digits in (2ABC)^4

    The product of two numbers is 7168 and their highest common factor is 16. How many pairs of numbers are possible such that the above condition is satisfied?

    Let us say that the numbers are x and y.
    Both of them are multiples of 16 as their HCF is 16
    So, we can say that they are 16a and 16b.

    Also, HCF (a,b) = 1. They have no factor in common otherwise the HCF of 16a and 16b won't be 16. In other words, a and b are coprime to each other.

    x * y = 7168
    => 16a * 16b = 7168
    => a * b = 28
    Now, we need to split 28 into coprime factors
    28 = 1 * 28 or 4 * 7
    => (x, y) = (16, 448) or (64, 112)

    So, there are two pairs that satisfy the given conditions.

    Note: I have considered only positive numbers for the calculation shown above. If you wish to consider negative numbers as well, the answer would double

    31 consecutive leaves were torn off. which of following could be the sum of 62 page numbers on these leaves?
    a) 1955
    b) 2201
    c) 2079
    d) none

    From the leaves that are torn off, the first page number should be odd.
    Let us assume it is 2a + 1.
    Since there are total of 62 pages which are torn off, the last page number will be 2a + 62
    Sum of 'n' terms in an Arithmetic Progression = n/2[First term + Last Term]
    => Sum of page numbers = 62/2 [2a + 1 + 2a + 62]
    => Sum of page numbers = 31(4a + 63)

    Now, let's look at the options
    Option (a) 1955 is ruled out because it is not divisible by 31

    Option (b) 2201
    => 31(4a + 63) = 2201
    => 4a + 63 = 2201/31 = 71
    => 4a = 71 - 63 = 8
    => a = 2
    => 2201 is a valid value

    Option (c) 2079 is ruled out because it is not divisible by 31

    How many times is a key of a typewriter pressed in order to type the first 299 natural numbers by pressing the space bar once between any two successive natural numbers?

    Number of 1 digit natural numbers {1, 2, 3 ... 9}= 9
    Keystrokes required to type them = 9*1 = 9

    Number of 2 digit natural numbers {10, 11, 12 ... 99} = 90
    Keystrokes required to type them = 90*2 = 180

    Number of 3 digit natural numbers {100, 101, 102 ... 299} = 200
    Keystrokes required to type them = 200*3 = 600

    Number of spaces between the numbers = 298
    Keystrokes required to type them = 298*1 = 298

    Total keystrokes required = 9 + 180 + 600 + 298 = 1087

    Which number is greater, 80^50 or 60^54?

    To solve this, it would help if you know Log(2) = 0.3010 and Log(3) = 0.4771
    If we have to compare a and b, we can compare Log(a) and Log(b).

    Log (60^54) = 54 Log (60) = 54 [Log (2) + Log(3) + Log(10)] = 54[0.3010 + 0.4771 + 1] = 54 * 1.771 = 95.634

    Log (80^50) = 50 Log(80) = 50 [Log(8) + Log(10)] = 50[3 Log(2) + 1] = 50[3*0.3010 + 1] = 50 * 1.9030 = 95.15

    So, Log (60^54) > Log (80^50) => 60^54 > 80^50

    How many 6 digit numbers can be made using digits 1, 2, 3, 4, 5, and 6 without repetition such that the hundred digit is greater than the ten digit, and the ten is greater than the one digit?

    Let us assume that the number that we have is abcdef
    The condition given is d > e > f

    Let us first select three digits for a, b, and c.
    From the given 6 digits (1,2,3,4,5,6), 3 digits for a,b, and c can be selected in 6C3 = 6!/3!3! = 20 ways.
    These three selected digits can be arranged on three positions of a, b, and c in 3! = 6 ways

    The remaining 3 digits are automatically selected for d, e, and f.
    So, we have 1 way of selecting them.

    The biggest digit will be allocated to 'd'.
    The second biggest digit will be allocated to 'e'.
    The third biggest digit will be allocated to 'f'
    So, we have 1 way of arranging them on the three positions of d, e, and f.

    Total ways = 20 * 6 * 1 * 1 = 120


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