Divisibility Rule Concepts For CAT  Hemant Malhotra

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
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 hereFind 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 0If 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) 8Number 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+BThe 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
question6If 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 5case1
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 9case2 last digt is 5
232x45 div by 3
so 2+3+2+x+4+5 =16+x should be div by 3so 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 above12AB31 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=18now div by 11
so 12+AB+31 shoud be div by 11
1+AB should be div by 11 so AB=1 or 10
but 10 is not possible bcz max value of A and B is 9 so difference couldn't be greater than 9so AB=1 , BA=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 17If 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
49+x8+1y = 12+xy div by 11
so xy could be 12 or 10 or 1
but 12 and 10 not possible so xy=1 so x=3 and y=2 possible value x=9 and y=8Let 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=8Consider 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 these999 = 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) 9divisible 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 000096+abc687+355
(428 + abc) mod 1001 = 0
so abc = 428
so sum of digit = 4 + 2 + 8 = 1415!=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=36834! = 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
000000+5ab643+609618+847140+604cd9+799232+295=2041cd9
so 2041cd9 mod 1001=0
so 39cd9 mod 1001=0
so c=0 and d=3Find 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 pairLet 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 viceversa.
In base 7 k = 7, k+1 = 8. 4 is the divisor of 8. The alternating sum of the digits of 123abc231bca312cab is 12+3a+bc+23+1b+ca+31+2c+ab = 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 = 86Bonus 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