Divisibility Rule Concepts For CAT - Hemant Malhotra


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    If number 234x32 is divisble by 8 find number of possible values of x ?

    Here number is divisble by 8 so last 3 digits will be divisible by 8 = > x32 is divisble by 8
    so x could be 0, 2, 4,6,8
    5 values of x is possible here

    Find sum and product of all possible values of x in number 2321256x where number is divisible by 4 ?

    Number is divisible by 4 so last two digits should be divisble by 4
    so 6x should be div by 4
    so possible values of x could be 0,4,8 so sum of all values of x is 12 and product is 0

    If a number 2341y is divisble by 5 find sum of all possible values of y ?

    Number is divisible by 5 so possible values are 0 and 5

    A number 12AB31 is divisible by 3, Which of the following cannot be the value of A+B ?

    a) 2  
    b) 3 
    c) 5 
    d) 8

    Number is divisible by 3 means sum of digits will be divisible by 3
    so 1+2+A+B+3+1 =7+A+B should be divisible by 3 so A+B could be 2,5,8,11,14 and so on means AP series with common difference as 3 and first term 2
    so here 3 is not a possible value of A+B

    The six digit number 24687X is divisible by 9 where X is a single digit whole number. Find X.

    Number is divisible by 9 so sum of digits should be divisible by 9
    2+4+6+8+7+X =27+x should be divisible by 9
    so x could bw 0 or 9 so we can't determine unique value here so Ans- can't be determined
    question6-If a number 232x4y is divisble by 15 then which of the following value of x+y is not possible
    approach = 232x4y is divisible by 15 so it will be div by 3 and 5
    for 5- last digit should be o or 5 so y could be 0 or 5

    case1-
    when y=0
    then 232x40 is div by 3 so sum of digit should be div by 3
    2+3+3+x+4+0=12+x should be div by 3 so x could be, 0,3 ,6 and 9

    case2- last digt is 5
    232x45 div by 3
    so 2+3+2+x+4+5 =16+x should be div by 3

    so x could be 2,5,8

    so these are the possible combination

    A number 12AB31 is divisible by 33, Which of the following cannot be the value of A+B ?

    a) 11
    b) 17
    c) 5
    d) 8
    e) None of the above

    12AB31 is div by 33
    33=3*11 so number is divisible by 3 and 11
    12AB31 is div by 3
    so 1+2+A+B+3+1 div by 3
    7+A+B is div by 3
    so A+B could be ,2,5,8,11,14,17(( bcz A and B can't be greater than 9 so max sum=18

    now div by 11
    so 1-2+A-B+3-1 shoud be div by 11
    1+A-B should be div by 11 so A-B=-1 or 10
    but 10 is not possible bcz max value of A and B is 9 so difference couldn't be greater than 9

    so A-B=-1 , B-A=1
    and A+B could be 2,5,8,11,14,17
    now use, B - A = 1 and A + B = 8
    = > B=9/2 which is not possible so only odd values of A+B is possible so possible values are 5,11 and 17

    If 49x81y is completely divisible by 66 then how many pairs of (x, y) values are possible ?

    div by 66 so div by 2,3 and 11
    div by 2 means last dgit y=0,2,4,6,8 possible
    now div by 3 so 22+x+y div by 3
    so 1+x+y div by 3 so possible values of x+y=2,5,8,11,14,17
    now div by 11
    4-9+x-8+1-y = -12+x-y div by 11
    so x-y could be 12 or -10 or 1
    but 12 and -10 not possible so x-y=1 so x=3 and y=2 possible value x=9 and y=8

    Let N = 111...111(73 times). When N is divided by 259, the remainder is a, and when N is divided by 32, the remainder is b. Then a+b is equal to
    A. 6
    B. 8
    C. 253
    D. None of these.

    259=37*7
    111111 ( 1 repeated 6 times ) is divisible by 259,37 and 7. now the given no. can be written as 111111000000... + 111111000000... + and so on..
    but the last '1' is left out. because there are 73 ,ones in the question hence the remainder is 1.
    as for 32... this is 2^5.. general rule is fro 2^n , check the last n digits , for remainder. for (2^1)2, last 1 digit is the deterministic factor..... checking 11111 for 259.. manually dividing 7 is the rem.
    so sum 7+1=8

    Consider a number formed by writing 2004 consecutive 9's (9999 … 2004 times). This number is not divisible by which of the following ?
    1) 7
    2) 13
    3) 37
    4) 101
    5) None of these

    999 = 37 * 27
    9999 = 99 * 101
    999999 = 999* 1001 = 999 * 11 * 91 = 999 * 11 * 13 * 7
    2004 is a multiple of 3, 4 and 6.
    The given number can be divided into groups of three 9's, four 9's or six 9's at a time.
    The given number is divisible by 7, 13, 37 and 101.

    17! = 355687abc096000, where a,b and c are digits, find the value of a+b+c
    a) 23
    b) 14
    c) 5
    d) 9

    divisible by 7 ,11,13 so divisible by 1001
    1001 = 10^3+1
    so make group of 3 and add with alternate sign from right side
    so 000-096+abc-687+355
    (-428 + abc) mod 1001 = 0
    so abc = 428
    so sum of digit = 4 + 2 + 8 = 14

    15!=1307674abc000, find digital sum of abc

    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

    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

    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 91
    1 pair

    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

    Bonus Funda

    n! has n digits when n= 1, 22, 23 and 24
    n! has less than n digits when 2 < =n < =21
    n! has more than n digits when n=0 or n > =25

     


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