Question Bank - Number Theory



  • (abc) base 9= (cba) base 11
    a * 81 + b * 9 + c = c * 121 + b * 11 + a
    a,b,c must be less than 9 and a,c can not be 0.
    a * 81 + b * 9 + c = c * 121 + b * 11 + a
    120c - 80a + 2b=0
    60c - 40a + b = 0
    c= 2, a=3, b=0 satisfies
    You can also check
    (302)9= (203)11
    But c=4 and a=6 will also satisfy.
    Also c= 6 and a=9, but a,b,c can not be 9, so ignore.
    So answer is CBD.



  • Q26) How many 2 digits number in base 9 gets reversed when expressed in base 11



  • (ab)9= (ba)11
    9a+b= 11b+a
    8a=10b
    a/b= 5/4
    a and b are single digit so only choice is 5,4.
    So 1 number.



  • Q27) Number of zeroes in 6! In base 6.



  • Number of trailing zeroes in base n: So in decimal system we used to check for highest power of 10, so in any base system we will check for the highest power of n.
    Example : Number of zeroes in 6! In base 2
    6/2 + 6/4 = 3 + 1 = 4

    Number of zeroes in 6! In base 6.
    6 = 3 * 2
    So check by 3 and 2. But we know 3 will be less so check directly by 3.
    6/3=2
    So 2 zeros



  • Q28) What is the remainder when 47^37^27 is divided by 11?



  • Q29) How many numbers between 0 and 500 are divisible by 3 or 5 or 7?



  • The question can be solved using the concept of set theory,
    Numbers that are divisible by 3, 5 and 7 can be represented as n(3), n(5) and n(7) respectively.
    Between 0 to 500,
    n(3) = 500/6= 166
    n(5) = 500/5 = 100
    n(7)= 500/7 = 71
    Total numbers = n(3) + n(5) +n(7)= 166+100+71 = 337

    Here, we have calculated numbers that are divisible by 3,5 and 7 individually.
    There are some numbers that have been repetitively counted, in order to eliminate them we have to find out the common numbers.

    Numbers that are divisible by 3 and 5 = n(3&5) = 500/ (LCM of 3 & 5) = 500/15 = 33
    Numbers that are divisible by 5 and 7 = n(5&7) = 500/ (LCM of 5 & 7) = 500/35 = 14
    Numbers that are divisible by 3 and 7 = n(3&7) = 500/ (LCM of 3 & 7) = 500/21 = 23
    Numbers that are divisible by 3,5 and 7 = n(3&5&7) = 500/(LCM of 3&5&7) = 500/ 105 = 4

    Formula of set theory can be used,
    n(A or B or C) = n(A)+n(B)+n(C)-n(A&B)-n(B&C)-n(A&C)+n(A&B&C)
    Substituting the values in the given formula, we obtain
    The actual numbers that are divisible by 3 or 5 or 7 = 337–33–14–23+4 = 271

    [solved by vivek subramanian]



  • Q30) M is a two digit number such that the product of the factorials of its individual digits is greater than the sum of the factorials of its individual digits. How many values of M exist?
    a) 56
    b) 64
    c) 63
    d) None of these



  • Approach: Numbers having 1/0 in them will not satisfy
    a! * b! > a! + b! as 1 * something is always less than 1 + something.
    So eliminate numbers having 1/0 in them
    11-91, 10-90, 12-19= 26 numbers
    Also, an exception is 22 where 2! + 2! = 2! * 2!
    Hence, total 27 numbers will not satisfy.

    So, 90-27=63 numbers will satisfy

    [solved by vinit sanghvi]


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