Free Online Preparation for CAT with Minglebox eCAT Prep. Cover basic concepts of Number Systems under Quantitative Aptitude for MBA Entrance Exam Preparation with Study material, solved examples and tests prepared by CAT coaching experts.
1 is the smallest Natural number, 0 is the smallest Whole number, and there is no largest or smallest Integer.
Even numbers are multiples of 2. Any even number can be written as 2n, where n is an integer. 0 is an even number.
Odd numbers are numbers which when divided by 2 leave a remainder of 1. Any odd number can be written as 2n +1, where n is an integer.
Factors of a given natural number say n, is another natural number say ,f if n is completely divisible by f. Ex factors of 18 are 1, 2,3,6,9,18 and number of factors is 6.
Highest factor of any natural number is number itself and lowest positive factor is 1. 1 is the factor of every natural number.
The number of factors of any natural number is finite.
A natural number which has exactly 2 factors is a prime number. Ex number 2 has factors 1 and 2 only. Similarly 3,5,7,11,13,17,19,23,29,31,37,41 etc . 1 is not a prime number.
A number n is a prime number if it is not divisible by any prime less than [√n] where [√n] is the largest natural number less than or equal to √n.
Any natural number n can be written in the form and in one unique way only n = p ^{ α} x q ^{ β} x r ^{ γ} ….., where p,q,r… are different primes and α,β,γ…. Are the powers of the prime number respectively.
Example
18 = 2 ^{ 1} x3 ^{ 2}. Here 2 and 3 are the prime numbers and 1 and 2 are the respective powers of the primes.
The number of factors of any natural number n, which can be factored as above is = (1+α)x(1+β)x(1+γ)…. etc. Thus number of factors of 18 is (1+1) x (1+2) = 6.
The number of odd factors will be given as follows. If the given number does not have any power of 2 then number of factors is same as number of odd factors. But if the number has any term like 2 ^{ α} in its factorization, as product of prime number powers , where α ≥ 1, then exclude the (1+α) term and calculate the number of odd factors as (1+β)x(1+γ)…. Etc. Thus number of odd factors of 18 is (1+2) = 3, viz... 1, 3 and 9.
The sum of all the factors is given by the expression ( p ^{ α1} 1)x( q ^{ β1} 1)x( r ^{ γ1} 1)/((p1)x(q1)x(r1)…..
Composite number is a number which has more than 2 factors. For example number 18 has 6 factors viz. 1,2,3,6,9,18.
Any whole number say m is divided by another natural number say n then there exists numbers q and r such that m = nxq + r. Where q is known as the quotient and r is known as the remainder. For any m ,n ? N(the set of natural numbers) q and r ? W(the set of whole numbers). Thus 0 ≤r
As discussed above the remainder obtained when any number is divided by say 5 then remainder is either 0 ,1,2,3,4 only. Therefore any number can be written as either as 5k , 5k+1 , 5k+2 , 5k+3 , 5k+4. Thus entire set of numbers has been split into 5 non overlapping sets.
HCF of any given set of numbers is a number which completely divides each number in the given set and the number is highest such number possible.
LCM of any given set of numbers is the smallest such number which is divisible by each number of the given set.
For any 2 given numbers HCFxLCM = Product of the 2 numbers.
HCF of fractions = HCF of numerators of all the given fractions/LCM of the denominators of all the fractions.
LCM of fractions = LCM of numerators of all the fractions/HCF of the denominators of all the fractions.
If the last digit of a number is even then number is divisible by 2
If the sum of the digits of a number is divisible by 3 then 9 then number is divisible by 3 and 9 respectively.
If the last 2 digits of the number are divisible by 4 then number is divisible by 4 and if last 3 digits is divisible by 8 then number is divisible by 8.
If the last digit of the number is 0 or 5 then number is divisible by 5
If the sum of digits of the number is divisible by 3 and the last digit is even then number is divisible by 6.
A given number is divisible by 7 if the number of tens in the original number  twice the units digits is divisible by 7 ex. to check whether 343 is divisible by 7 or not , Thus twice the units digit is 2x3=6 and number of tens in the number is 34. Therefore 343 is divisible by 7 if 34 6 is divisible by 7 i.e. 28 is divisible by 7.
A number is divisible by 11 if the difference of the sum of digits occurring the even numbered places and the sum of digits occurring in the odd number of places is divisible by 11.
Rules of Indices
b ^{ l} xb ^{ m} xb ^{ n} …..= b ^{ l+m+p…..}
(p ^{ m} ) ^{ n} = p ^{ mn}
If bases are same then powers are also same. i.e. if a ^{ m} = a ^{ p} and if a ≠1 or 0 then it implies m = p.
The last digit of the square of any number cannot be 2,3,7 or 8.
Any perfect square is of the form either 4k or 4k1 or 4k+1.
Product of any number of even numbers is even and any number of odd numbers is odd.
The product of any n consecutive natural numbers is divisible by n!.
Useful Algebraic identities
(a ^{ n} +b ^{ n} ) is divisible by (a+b) for all odd values of n.
(a ^{ n}  b ^{ n} ) is divisible by both (a+b) and (ab) for even values of n.
(a ^{ n}  b ^{ n} ) is divisible by (ab) for all values of n (both odd and even).
Sum of first n natural numbers = n(n+1)/2.
Sum of squares of first n natural numbers = n(n+1)(2n+1)/6
Sum of the cubes of first n natural numbers is given by {nx(n+1)/2} ^{ 2}.
Units digit of any power of a number (Cyclicity)
If we consider the units digit of the powers of 2 i.e. 2 ^{ x} for different values of x then we find the unit’s digit is 2,4,6,8,2,4,6,8,2,4,6,8…. for x= 1,2,3,4,5,6,7,8,9,10,11,12…. Similar patterns exist for the unit’s digit of other numbers. The results are summarized in the table below.

Unit's digit of a^x, k is an natural number 
Number "a" 
x=4k+1 
x=4k+2 
x=4k+3 
x=4k 
1 
1 
1 
1 
1 
2 
2 
4 
6 
8 
3 
3 
9 
7 
1 
4 
4 
6 
4 
6 
5 
5 
5 
5 
5 
6 
6 
6 
6 
6 
7 
7 
9 
3 
1 
8 
8 
4 
2 
6 
9 
9 
1 
9 
1 
QUESTION
The Rightmost non zero digit of 5670 ^{ 5670} is
(a) 7 (b) 1 (c) 3 (d) 9
SOLUTION
Here we use rule of cyclicity to solve this problem. The first non zero digit to the left of 0 is 7 hence the rightmost non zero digit will be same as the units digit of 7 ^{ 5670}. Now next we need to find out the form of 5670. Dividing 5670 by 4 we get 2 as remainder. Hence 5670 is of the form 4k +2. Therefore the unit’s digit is 9. Therefore answer is d
QUESTION
The numbers 13409 and 16760 when divided by a 4 digit integer n leave the same remainder then the value of n is
(a) 1127 (b)1117 (c) 1357 (d)1547
SOLUTION
Here 13409 and 16760 on division by n leave the same remainder hence can be written as nxm + r and can be written nxp + r. Therefore subtracting the 2 equations we get nx(mn) =3351 = 1117x3 But n is 4 digit number hence is n is 1117. Hence answer is b.
QUESTION
If N = 82 ^{ 3} – 62 ^{ 3} 20 ^{ 3} then N is divisible by
(a) 41 and 31 (b) 13 and 67 (c) 17 and 7 (d) None of these
SOLUTION
Using the identity a ^{ 3} + b ^{ 3} + c ^{ 3} 3abc = (a+ b+ c)x(a ^{ 2} + b ^{ 2} +c ^{ 2} – ab  ac –ac). We find that if a + b + c = 0. Then a ^{ 3} + b ^{ 3} + c ^{ 3} = 3abc. Therefore a ^{ 3} + b ^{ 3} + c ^{ 3} is divisible by a , b , c and 3. Now a = 82 , b = 62 and c = 20. Therefore N is divisible by 3 , 82 , 62 and 20. But 82 is divisible by 41 and 31. Hence answer is a.
QUESTION
Let X be a set of positive integers such that every element a of S satisfies the condition (i) 1100 ≤ n ≤ 1300 (ii) every digit in n is odd
Then how many elements of X are divisible by 3?
(a) 10 (b) 9 (c) 16 (d) 13
SOLUTION
We know that any number is divisible by 3 if the sum of the digits of the number is divisible by 3. Now between 1200 and 1299(both inclusive) we find that all the numbers have one digit viz. 2(the hundreds digit) which is even, hence there do not exists any number between 1200 and 1299 with the given conditions. Also 1300 is not divisible by 3. Therefore the required number will lie between 1100 and 1199. Between 1100 and 1199 all the numbers will have first 2 digit as 1 Therefore if we consider any number between 1100 and 1199 it will be of the form 11ab , where a, b are digits. Sum of digits will be given by 1 +1 +a +b i.e. 2+a+b. Now a, b are digits therefore maximum value of a +b can be 18 hence max value of 2+a+b can be 20. Therefore for 11ab to be divisible by 3 value of 2+a + b can be either 18 , 15, 12, 9 , 6 or 3. Also a,b has to be odd ,therefore a +b will be even. Hence 2+ a +b will be even. Hence 2 + a +b can assume values which are even. Therefore 2 + a+ b can be either 18 or 12 or 6 only. Which => a+b can be 16, 10, and 4 only. Solving a +b =16 for odd a, b we get a=9, 7 and b = 7, 9 respectively. Similarly solving a +b = 12 we get a = 9, 7, 5,3 and b = 3,5,7,9 respectively and solving a + b =6 we get a = 5 , 3 ,1 and b = 1 ,3,5 respectively so total number of solution is 9.
QUESTION
Four bells toll at intervals of 15, 18, 24 and 32 minutes respectively. At a certain time , they begin to toll together after what , least time interval of time(in mins) will they toll together again?
(a) 1440 (b) 900 (c) 1660 (d) 1335
SOLUTION
The first bell tolls at after every 15 minutes i.e. at 15 minutes , 30 minutes , 45 minutes etc. Similarly the second bell will toll at 18 minutes , 36 minutes etc, Thus we observe that all the 4 bells will toll together at a time which is common multiple of 15 , 18 , 24 and 32. Also as we are looking for least such interval hence the required time is the LCM of 15 , 18 , 24 and 32 which is 1440 mins
marihal : THANKS
vikas77065 : keep up the great work!!
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manish48464 : Its nice plateform for prepratiom of CAT. I would like to thanks to all team & members of this Plateform.
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utsav7898 : thanx
kirti73125 : Are these tests getting launched or just d fake ones..... because i tried taking all the tests and none of them actually started.... it says "press Begin" button to take test....but ther's nothing like that.
girish51604 : Hey content is really nice..
but I found one simple mistake.
I.e in table for powers 4k, 4k+1...
The sequence for 2 is 2,4,8,6,2,4,8,6.... not 2,4,6,8.
nitti54202 : tanpreetkaur7: its gud to be a part of minglebox helps a lot
bhanu01881 : the material was nic but how can i download??????
gaurav1025 : its good actually some of the problems which r easy but we hesitate to ask to the tutor bt here it is easy to get ....
shambhuaharansharma : @Rahul:
Answer will be 172..
Max HCF= 1680
to get 4 remdr...the number should be 1684...
so min numb u need to subtrsct is: 1856172=1684
answer: 172
ramasundaram52424 : Very useful!Thanks a lot!
ravi26952 : gud thanx
kumar6595 : gd ques
riya23704 : The questions are of easy level. I doubt if these are CAT level nowadays. Its just my humble opinion. I liked the questions on bases though.
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jyotsna98484 : The sum of all the factors is given by the expression ( p α1 1)x( q β1 1)x( r γ1 1)/((p1)x(q1)x(r1)…..
is not correct.
The correct expression is
( p α+1 1)x( q β+1 1)x( r γ+1 1)/((p1)x(q1)x(r1)…..
debojyoti85192 : in the solved qsn no. 3....there is a mistake.plz rectiofy tht.......hw can 82 uis divisble by 31?
paritala54873 : these are good.but i think is it enough to go furthur?
paritala54873 : excellent keep it up.
amit2931 : hello hw cn i launch d practice test?
sujith4865 : Really, these are good! students who are not able to take coaching for cat, this minglebox will be their good alternative
vivek32027 : can u please help me with this ques
find the value of n which divides 77!/5^n..how to solve such pblms. please tell me amothod to solve such pblms
ratan1349 : i want coaching for mba cat
vijay9219 : Hii sharda.. see if u want to find whether a number 'x' is prime or not then u simply find out the square root of the number and round it off to its next value, say 'y' and then just divide the number 'x' till all the prime numbers upto 'y'. As simple as that.
anuroop7637 : If (x^2)(y^3)z = (p^2)(q^3)r, is y = q?
1) x^2 = r and p^4 = z^2
2) x,y and z are distinct primes and so are p,q and r.
In this we can find the answer from 1st op and not from 2nd op bt here it is given that we can find it from 2nd op and not from 1st op can ne1 explain????
abhishek9230 : plz chek the table for cyclicity all the values are correct except for 4k+3 and 4k for unit digit 2
sundersriram : thanx
ankit7714 : In CYCLICITY, if unit digit is 2 then 4k+3 will be 8 not 6 and 4k will be 6 not 8.
Indeed, It is really helpful for any one. Thank you for those who developed this marvelous web..........
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shradha3441 : anyone who can explain dis statement more briefly then wud plz. do for me!!!
"A number n is a prime number if it is not divisible by any prime less than [√n] where [√n] is the largest natural number less than or equal to √n."
confused?????
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hemant102 : more studu material and important concepts is reqd like square test etc.
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akash3876 : 2 questions were answered incorrect though the explanation below them was fine in the First test. Please correct.
sahitya6000 : In the section where units digit is considered for 2^x, the pattern should be 2, 4, 8, 6 instead of 2, 4, 6, 8......
Sorry to point out...but thought that others might get confused as well like me...
akhillar : thanks for everything.its amazing.
rahul124 : it cool
vk5613 : Hi after i click the launch test i go to another window with instructions but dont see any start test...what do you think the problem could be?
robinitsme4me : What is the greatest fivedigit number which is a perfect square?
a) 99225
b) 98596
c) 97969
d) 99856
answer is d,,,,,ur site is sayng its 97969....
d = 316^2
madhurendra9785 : Thanks to all the team of mingle box 4 this amazing site for freshers help who are going to preparing for CAT.
Good Job.
Well Done.
shilpi3400 : gud job!!
bu please give the explanations and answers alittle more concise and ofcourse correct.
rest its fab to practise.
thanx minglebox..
anklush : good and need some more practice qustn
pusarla1734 : please check the last but one answer...i think it is 10..a+b must be either 4,10,16..but u considered a+b as 12,6 so the total sum will be 14 and 8 which is nt divisible by 3..please check the answer
lavin8553 : formula for solved xample is wrong its abbcca
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sumit2530 : Awesome materials...
piyush6097 : hello!
77^77/ 19
(76+1)^77/19
(76k)^77+1^77/19 hence(76k)^77 is divisible by 19 so 1^77= 1 is the remainder
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ravi4804 : hiii..can any body help me in understanding d topic of cyclicity....and d model sums on LCM
rahul7688 : its very helpfull to preperation for cat
tulasi3490 : IN THE REFRESHER TEST THE SOLUTIONS FOR 4 AND 10 ARE WRONGLY TYPED. FOR 4) IT IS 174 AND FOR 10) IT IS A & B.
sourav3194 : The least number that must be subtracted from 1856 so that the remainder when divided by 7,12,16 is 4?
a) 137
b) 174
c) 140
d) 172
Explanation
1856  x = max*LCM + 2 LCM = 336 .'. x = 1856336*52 = 174
can u justify the solution provided as i dnt find dat wen 1682(1856174) is divided by 7 leaves remainder 4.
The answer should be 172...Pls Check
sakshi9666 : its pretty cool yaar.....but i think sum material is still missin
balamuruganb : Awesome materials...
atul2777 : see buddy u can rit 77^77/19 as (1+76)^77/19 okay this will leave 1^77 ="1" as a remainderas 76 is divisible by 19 .
any further queries are welcomed!
shiwani6525 : its really so helpful
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yadavrajesh : its really excellent, very helpful
atul2777 : very good
aseem5527 : hi how to solve
Q. 77^77/ 19 ?
we have to find remainder ?
can anyone tell me how to solve?
thanks in advance.
aseem5527 : very nice and good test
james5930 : The provided solutions of two questions of Refresher Test1 are wrong.Administrator,please correct them.One is "The least number that must be subtracted from 1856 so that the remainder when divided by 7,12,16 is 4?".The correct answer is 172 not 174.The answer of question no.10 is also wrong.
kapil2213 : its excellent and very good
nikkipandey : its gr8.......
anirudhmr : Thank you rahul as you pointed out the remainder obtained when 174 is subtracted is 2 and not 4, So if we subtract 172 we get the right answer.
So the correct answer is 172 i.e d
p.s keep posting !
tikkiwal : material has a few mistakes,, so plz try 2 remove d errors..
tikkiwal : QUESTION
The Rightmost non zero digit of 5670 5670 is
(a) 7 (b) 1 (c) 3 (d) 9
SOLUTION
Here we use rule of cyclicity to solve this problem. The first non zero digit to the left of 0 is 7 hence the rightmost non zero digit will be same as the units digit of 7 5670. Now next we need to find out the form of 5670. Dividing 5670 by 4 we get 2 as remainder. Hence 5670 is of the form 4k +2. Therefore the unit’s digit is 9.Therefore answer is d
it is of the form of 4k+3
so answer must be 3 that is C
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rahul9060 : here is a question 4m one of ur review tests
The least number that must be subtracted from 1856 so that the remainder when divided by 7,12,16 is 4?
a) 137
b) 174
c) 140
d) 172
Explanation
1856  x = max*LCM + 2 LCM = 336 .'. x = 1856336*52 = 174
can u justify the solution provided as i dnt find dat wen 1682(1856174) is divided by 7 leaves remainder 4.