Factors and Multiples play a crucial role while studying Number system for CAT especially while playing with numbers and getting results out of it. It will also be helpful while solving DILR sets filled with numbers.
Product
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The result of multiplying two or more numbers together is known as the product of these numbers.
Ex: 3 × 4 = 12 . Here 4 and 3 are the factors of 12.
12 is called as the multiple of 3 or multiple of 4.
A factor (or divisor) of a number is a number that perfectly splits that number, and that number is called a multiple of that number.
15 is exactly divisible by 1, 3, 5, 15 so 1, 3, 5, 15 are called as the factors of 15 while 15 is called as the multiple of these factors
Rules of factors and multiples
- 1 is a factor of every number.
- Every number is a factor of itself.
- Every number, except 1 has atleast two factors viz., 1 and itself.
- Every factor of a number is less than or equal to that number.
- Every multiple of a number is greater than or equal to itself.
- Every number has infinite number of its multiples.
- Every number is a multiple of itself.
Prime Factorisation
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If a natural number is written as the sum of prime numbers, which are called factors, then the number's factorization is called its prime factorization.
Ex: Prime Factor of 48
Number of Prime Factors of a Composite Number
Let N be a composite number with prime factors a, b, c, d, etc., and let p,q,r,s, etc. be the indices (or powers) of a,b,c,d, etc., then the number of total prime factors (or divisors) of N is = p + q + r + s + ……
Number of All the Factors of a Composite Number
Let there be a composite number N and its prime factors be a, b, c, d, ...etc. and p, q, r, s, ... etc, be the indices (or powers) of the a, b, c, d, ...etc, respectively. That is if N can be expressed as N =ap.bq .cr .ds..., then the number of total divisors or,
factors of N is ( p +1) (q +1) (r +1) (s +1) ...
Number of Odd Factors of a Composite Number
Number of odd factors of a number N can be expressed as (pa . pb . pc ....) . (ex)
(where, p1 , p2 , p3 , ... are the odd prime factors and e is the even prime factor.)
Then the total number of odd factors = ( a + 1) ( b + 1) ( c + 1) . . .
Number of Even Factors of a Composite Number
Number of even factors of a number = (Total number of factors of the given number – Total number of odd factors)
Sum of Factors of a Composite Number
Let N be the composite number and a, b, c, d.. be its prime factors and p, q, r, s be the indices (or powers) of a, b, c, d. That is if N can be expressed as
N =ap .bq .cr .ds...
then the sum of all the divisors (or factors) of N
\(\frac{(a^{p+1}-1) (b^{q+1}-1) (c^{r+1}-1) (d^{s+1}-1)}{(a-1)(b-1)(c-1)(d-1)} .....\)
Product of Factors of a Composite number
the product of factors of composite number N = N n/2 , where n is the total number of factors of N .
Number of Ways of Expressing a Composite Number as a Product of two Factors
the number of ways of expressing a composite no. as a product of two factors = \(\frac{1}{2}\) x the no. of total factors
the no. of ways of expressing N as a product of Two distinct factors then we do not consider 1 way (i.e., N = \(\sqrt {N}\) x \(\sqrt {N} \)) then
Number of ways = \(\frac{(Number of Factors \ - \ 1)}{2}\)
Number of Co-Primes of A Composite Number N
For a composite number N = a p × bq × cr .., the number of co-primes of N is given by
F(N) = \(N(1- \frac{1}{a}) (1- \frac{1}{b}) (1- \frac{1}{c}) ......\)
a,b,c …are prime numbers
Sum of All The Co-Primes of a Composite Number N
If composite number N = a p × bq × cr ...., then
The sum of all the co-primes of N is given by
\(\frac{N}{2} [F(N)]= \frac{N}{2}[N(1- \frac{1}{a})(1- \frac{1}{b})(1-\frac{1}{c})......]\)
a,b,c are prime numbers
Number of Ways of Expressing a Composite Number As A Product of Co-Prime Factors
There are 2n-1 ways to write a composite number N as the product of co-prime factors, where n is the number of unique prime factors of N.
HCF
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HCF of two or more than two numbers is the largest number that can exactly divide all these numbers without leaving a remainder.
There are two methods of finding the HCF.
- Factor Method
- Division Method
Factor Method
In this method, the numbers are first resolved into their prime factors, then the product of their common factors is calculated. This product is referred to as the HCF of the provided numbers.
Division Method
Consider the two smallest numbers, then divide the larger one by the smaller one, then divide this divisor by the remainder, and so on, until the remainder equals zero. If there are more than two numbers for which the HCF is to be determined, the process continues by dividing the third lowest number by the last divisor obtained.
Be‘zout’s Identity
If ‘H’ be the HCF of any two positive integers a and b then there exists unique integers p and q such that
H = ap + bq
LCM
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if we consider any 3 numbers say, 3, 5 and 6 then the multiples of each of the 3, 5, 6 are as follows:
3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48, 51, 54, 57, 60, ...
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, ...
6, 12, 18, 24, 30, 36, 42, 48, 54, 60,...
So, we can see that the common multiples of 3, 5 and 6 are 30, 60, 90, 120, ... etc.
30 is the least common multiple.
LCM with Remainders
Case 1. When the remainders are same for all the divisors.
Case 2. When the remainders for different divisors are distinct, but the difference between the divisors and remainders remains constant, this is known as a constant remainder.
Case 3. When neither the divisors nor the differences between the divisors and remainders are the same and remain constant.
CASE – 1
What is the least possible number which when divided by 24, 32 or 42 in each case it leaves the remainder 5?
LCM of 24, 32 and 42 is divisible by the given numbers. So the required number
= (LCM of 24, 32, 42) + (5) = 672 + 5 = 677 Hence such a least possible number is 677.
how many numbers are possible between 666 and 8888?
number is 672m + 5, where m = 1, 2, 3, ...
So, the first no. = 672 × 1 + 5 = 677 and the highest possible number in the given range = 672 × 13 + 5 = 8736 + 5 = 8741
Thus the total numbers between 666 and 8888 are 13.
Case – 2
What is the smallest number that, when divided by 2, 3, 4, 5, 6, results in remainders of 1, 2, 3, 4, 5?
Since the difference is same as
(2 − 1) = ( 3 − 2) = (4 − 3) =(5 − 4) (6 − 5) = 1
Hence the required number = (LCM of 2, 3, 4, 5, 6) – 1 = 60 − 1 = 59
what is the least possible 3 digit number which is divisible by 11?
the number is (60m – 1), where m = 1, 2, 3, ....
but the number ( 60m − 1) should also be divisible by 11 hence at m = 9 the number becomes 539 which is also divisible by 11.
Thus the required number = 539.
Case – 3
How many numbers lie between 11 and 1111 which when divided by 9 leave a remainder of 6 and when d Let the possible number be N then it can be expressed
Divide by 21 leave a remainder of 12?
N = 9k+ 6 andN = 21l+12
9k+ 6 = 21l+12
9k−21l = 6
k = \(\frac{7l+2}{3}\)
the min. possible value of l such that the value of k is an integer or in other words numerator (i.e., 7l + 2) will be divisible by 3.
CAT Previous Year questions
Ques 1: The values of the numbers 22004 and 52004 are written one after another. How many digits are there in all? (CAT 2011)
- 4008
- 2003
- 2004
- None of these
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Ans: (D)
The given numbers are 22004 and 52004.
Let a = 2004
Total number of digits when 21 and 51 are written side by side (25) = (1 + 1)
Total number of digits when 22 and 52 are written one after another (425) = (2 + 1)
For 2 and 5 = (3 + 1).
∴Total number of digits when 22004 and 52004 is 2004 + 1 = 2005
Ques 2: Auto fare in Bombay is Rs. 2.40 for the first 1 km, Rs. 2.00 per km for the next 4 km and Rs. 1.20 for each additional km thereafter. Find the fare in rupees for k km (k ≥ 5). (CAT 2011)
- 2.4k + 1.2(2k − 3)
- 10.4 + 1.2(k − 5)
- 2.4 + 2(k−3)+1.2(k−5)
- 10.4 + 1.2(k−4)
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Ans: (D)
for k ≥ 5,
2.4 + 4 × 2 + 1.2(k − 5)
= 10.4 + 1.2(k − 5)
Ques 3: The largest number amongst the following that will perfectly divide 101100 − 1 is (CAT 2010)
- 100
- 10000
- 100100
- 100000
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Ans:(B)
By trail and error or by back substituting answers in the choices given.
1012 =10201
1012 − 1 = 10200
This is divisible by 100.
For
1013 −1 =1030301 −1 =1030300
(1019 − 1) will be divisible by 100.
(10110-1) to (10199-1) will be divisible by 1000
Therefore (101100-1) will be divisible by 10000
Ques 4: Rohan and Sohan take a vacation at their grandparents’ house. During the vacation, they do any activity together. They either played Tennis in the evening or practiced Yoga in the morning, ensuring that they do not undertake both the activities on any single day. There were some days when they did nothing. Out of the days that they stayed at their grandparents’ house, they involved in one of the two activities on 22 days. However, their grandmother while sending an end of vacation report to their parents stated that they did not do anything on 24 mornings and they did nothing on 12 evenings. How long was their vacation? (CAT 2010)
- 36 days
- 14 days
- 29 days
- Cannot be determined
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Ans: (C)
Let the number of days that they holidayed be equal to T.
they practiced Yoga on (T − 24) mornings. They played tennis on (T − 12) evenings.
As they did not do both the activities together on any single day, Days on which they had any activity = Number of days they practiced Yoga + Number of days they played tennis i.e.,
22 =T −24 + T −12
⇒ 22+24+12=2T
⇒58=2T
⇒T =29
Ques 5: A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then, M equals (CAT 2010)
- 31
- 63
- 75
- 91
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Ans: (D)
Since, the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91.
Among these only 31 and 91 are a part of the answer choices.
Among these, (31)10 , (11111)2 , (1011)3 , (111)5
Thus, all three forms have leading digit 1.
Thus , answer is 91.
Ques 6: A manager is not used to work in the decimal system. She says that there are 100 employees in the office of which 24 are males and 32 are females. Which number system does the manager use? (CAT 2009)
- 4
- 6
- 8
- 16
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Ans: (B)
Let the required number system = N, then
(100)N = (24)N + (32)N
N2 =2N +4+3N +2=5N +6
N 2 − 5N − 6 = 0
⇒ (N +1)(N −6) = 0
So, N = 6 as − 1 is not possible.
Ques 7: Theintegers1,2,...,40 are written on a black board. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b − 1 is written. What will be the number left on the board at the end? (CAT 2008)
- 820
- 821
- 781
- 819
- 780
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Ans: (C)
According to the question, if two numbers, say a and b, are wiped and replaced by a new number, a + b- 1, the number of integers will decrease by one with each repeat, until there is only one number left after the last repetition.
the sum of integers from 1 to 40 = \(\frac{n(n+1)}{2}\) = \(\frac{40\times41}{2}\) = 820
the total number of values in the first, second, third, etc. repeats will be 819, 818, 817, etc. So, after 39 steps, there will only be one number left, which is 820 39 = 781.
Ques 8: What are the last two digits of 72008 ? (CAT 2008)
- 21
- 61
- 01
- 41
- 81
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Ans: (C)
The last two digits of number the expansion of (7)4 = 01(2401) and if the power of 7 is any multiple of 4 the last two digits will not change,
i.e., (7)4 = 2401 ⇒ 01
(7)8 =5764801⇒01
Since, power of 7, i.e. 2008 is a multiple of 4, the last two digits of (7)2008 will be 01
Ques 9: The number of common terms in the two sequences 17, 21, 25, ...,417 and 16, 21, 26, ...,466 is (CAT 2008)
- 78
- 19
- 20
- 77
- 22
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Ans: (C)
Both the sequences (17,21,25, ... ) and (16,21,26,...) are arithmetic progression with a common difference of 4 and 5, respectively.
In both sets, the first term that is the same is 21. So, you can make a new mathematical sequence with a common difference of the LCM of (4, 5), which is 20. This new sequence will have the same terms as both of the old ones.
∴New sequence will be 21, 41, 61, ..., 401.
nth term = a + (n −1)d
401 = 21+(n-1)20
(n – 1) = 401-21/2
n = 20
Ques 10: How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0,1, 2, 3 and 4, if repetition of digits is allowed? (CAT 2008)
- 499
- 500
- 375
- 376
- 501
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Ans: (D)
The number required is greater than 999 and less than and equal to 4000. Now out off four digits 0, 1, 2, 3 and 4. To form a number greater than 999 and less than 4000.
The digit at thousands place can be selected in
3 ways. (∴0 and 4 can not be taken)
The digit at hundreds place can be selected in 5 ways. The digit at tens place can be selected in 5 ways.
The digit at unit place can be selected in 5 ways.
∴ Total required number of ways = 3 ×5 ×5 ×5 = 375 Since, 4000 is also one of the required number.
Therefore, total number of ways = 375 + 1 = 376
Ques 11: What is the number of distinct terms in the expansion of (a + b + c)20 ? (CAT 2008)
- 231
- 253
- 242
- 210
- 228
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Ans:
Ques 12: Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers? (CAT 2008)
- 1 \(\leq\) m \(\leq\)3
- 4 \(\leq\) m \(\leq\) 6
- 7 \(\leq\)m \(\leq\) 9
- 10 \(\leq\)m \(\leq\)12
- 13 \(\leq\)m \(\leq\)15
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Ans: (A)
Let the three consecutive positive integers be (n−1), n and (n+1).
⇒ n−1+n2 +(n+1)2 =(3n)2
⇒ n3 +4n2 +4n=9n2
⇒n2 −5n+4=0
⇒ n = 1 or n = 4
Since, the three integers are positive, the value of ‘n’ cannot be equal to 1, therefore the value of n = 4 or m = n − 1 = 3. Hence, three consecutive integers are 3, 4 and 5.
Hence, option (A) is the correct choice.
Ques 13: Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares? (CAT 2007)
- 2
- 4
- 0
- 1
- 3
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Ans:
If the first two digits of a four-digit number are the same and the last two digits are also the same, the number will have the form 11 (100a + b),
which means it will be a multiple of 11, like 1122, 3366, 2244, etc.
Let the number needed be aabb. Since aabb is a perfect square, a = 7 and b = 4 are the only numbers for a and b that meet the above condition.
So, 7744 is a square number.
Ques:14: Ten years ago, the ages of the members of a joint family of eight people added up to 231 yr. Three years later, one member died at the age of 60 yr and a child was born during the same year. After another three years, one more member died, again at 60 and a child was born during the same year. The current average age of this eight-member joint family is nearest to (CAT 2007)
- 22 yr
- 21 yr
- 25 yr
- 24 yr
- 23 yr
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Ans: (D)
Total age of eight people 10 yrs ago = 231yr
Total age of eight people 7 yrs ago
=231+8×3−60+0 =195
Total age of eight people 4 yrs ago
=195+3×8−60+0 =159
Current total age of eight people
=159+4×8 =191yr
Average age = 191/8 = 24 yrs. (Approx.)
Ques15: Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos? (CAT 2007)
- 16
- 18
- 15
- 19
- 17
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Ans: (B)
Let the number of currency of 1 miso, 10 misos and 50 misos be x, y and z, respectively.
Then, x + 10y + 50z = 107.
Now, possible values of z = 0, 1, 2.
If z = 0, then
x + 10y = 107.
Now, number of pairs of values of x and y that satisfy the above equation are 11.
These pairs (7,10), (17, 9), ..., (107, 0).
If z = 1, then x + 10 y = 57.
For this number of pairs of values of x and y is 6.
(7, 5), (17, 4), (27, 3), ... , (57, 0) if z = 2, then
x + 10 y = 7.
There is only one such pair of x and y, (7, 0) which satisfy the equation.
Therefore, total number of ways =11+6+1 =18.
How to approach Factors and Multiples concepts on CAT
- Go through all the formulae given in the factors section.
- Tricks for finding LCM and HCF for finding factors and multiples should be remembered
- Thorough with the Basic Number system questions will be an added advantage while solving tricky and tough questions in CAT.
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