It is one of the important chapters in the case of Number system and Algebra. It lays the foundation stone for both the sections. Everything in this section is of utmost importance. There are lot of rules to be used across Quantitative section in all the management entrance examinations.
Indices and Surds
[Click Here for Previous Year Questions]
Laws of Indices
If m and n are positive integers
- am x an = a m+n
- \(\frac{a^m}{a^n}\) = a m-n , (a \(\neq\)0)
- (ab)n = an . bn
- \((\frac{a}{b})^n = \frac{a^n}{b^n}\) , (b \(\neq\) 0)
- (am)n = am.n
- a0 = 1 , (a \(\neq\)0)
Some Important Results
- If ax =k, then a=(k)1/x
- If a1/x =k, then a=kx
- If ax = by, then a = (b)y/x and b = (a)x/ y
- If a x = a y , then x = y, where a \(\neq\) 0, 1
- a−n =1/an and an =1/a−n
- abc \(\neq\) (ab)c ; b \(\neq\) c
Surds
When a root of a rational number (i.e., quantities of the type \( \sqrt[n]{a}\), a being a rational number) can not be exactly obtained, then this root is called a surd.
Properties of Surds
- A quadratic surd cannot be equal to the sum or difference of a rational number and a quadratic surd e.g., \(\sqrt{a} + \sqrt{b}\) ≠ cor \(\sqrt {a} - \sqrt {b}\) ≠ c,where \(\sqrt a\) , \(\sqrt c\) are quadratic surds and b is a rational number.
- The product and quotient of two dissimilar quadratic Surds cannot be rational
- If \(\sqrt a + \sqrt b\) = c + \(\sqrt d\) or a – \(\sqrt b\) = c− \(\sqrt d\) then a = c and b = d.
- If a ± \(\sqrt b\) =0 then a=b=0
- If a + \(\sqrt b\) = c + \(\sqrt d\), then a − \(\sqrt b\) = c - \(\sqrt d\)
- If \(\sqrt {a+ \sqrt b}\) = \(\sqrt c+ \sqrt d\) + then \(\sqrt {a-\sqrt b}\) = \(\sqrt c - \sqrt d\)
Factorials
[Click Here for Previous Year Questions]
The product of n consecutive natural numbers (or positive integers) starting from 1 to n is called as the factorial ‘n’.
- Factorial n is written as ‘‘n! ’’.
- 0! = 1and 1! = 1
Properties
- n! is always an even number if n ≥ 2
- n! always ends with zero if n ≥ 5
The number of zeros at the end of the product depends upon 2 × 5, but the condition is that
- 2k × 5l gives k number of zeros if k < l
- 2k × 5l gives l number of zeros if l < k
Divisibility of a Factorial Number by the Largest Power of Any Number
Suppose we have to find the highest power of k that can exactly divided n!, we divide n by k, n by k2, n by k3...and so
on till we get \([\frac{n}{k^{-x}}]\) equal to 1 (where, [P ] means the greatest integer less than or equal to P) and then add up as
\([\frac{n}{k^1}] + [\frac{n}{k^2}] + [\frac{n}{k^3}] + [\frac{n}{k^4}]......... + [\frac{n}{k^n}] \)
Involution
[Click Here for Previous Year Questions]
a . a . a... (n times) = a n . So the process of multiplication of a number several times by itself is known as INVOLUTION.
Algebraic expressions that are often used while solving questions are:-
- (a+b)2 =a2 +b2 +2ab=(a+b)(a+b)
- (a−b)2 =a2 +b2 −2ab=(a−b)(a−b)
- (a+b+c)2 =a2 +b2 +c2 +2(ab+bc+ca) = (a + b + c) (a + b + c)
- (a+b)3 =a3 +b3 +3ab(a+b)
- (a−b)3 =a3 −b3 −3ab(a−b)
- (a+b+c)3 =a3 +b3 +c3 +3(a+b)(b+c)(c+a)
- (a + b)2 = (a − b)2 + 4ab
- (a − b)2 = (a + b)2 − 4ab
- a2 −b2 =(a+b)(a−b)
- a3 +b3 =(a+b)(a2 +b2 −ab)
- a 3 − b3 = (a − b) (a 2 + b2 + ab)
- a 3 + b3 + c3 − 3abc = (a + b + c) (a2 + b2 + c2 −ab − bc − ac)
Finding digits the remainder or divisibility of a number
- an + bn is never divisible by (a − b), where n is odd
- an + bn is never divisible by (a − b)
- an + bn is not divisible by (a + b), when n is even.
- an − bn is always divisible by (a − b)
- an − bn is divisible by (a + b) only, when n is even
- an − bn is not divisible by (a + b), when n is odd.
Concepts of Remainder
[Click Here for Previous Year Questions]
- Let us assume that when a1 , a2 , a3 , ..., an are individually divided by d , the respective remainders are R1, R2, R3, ..., Rn . Now, if we divide (a1 + a2 + a3 +K + an ) by d, we get the same remainder as when we get by dividing (R1 + R2 + R3 + K + Rn) by d.
- When ‘a’ is divided by ‘d’ the remainder is R and when ‘a1’ and ‘a2’ are divided by ‘d’ the remainders are R1 and R2, then the remainder R will be equal to the difference of remainders R1 and R2 if a1 − a2 = a.
Previous year CAT Questions
Ques 1: Let S be a set of positive integers such that every element n of S satisfies the conditions
- 1000 ≤ n ≤ 1200
- every digit in n is odd
Then, how many elements of S are divisible by 3? (CAT 2005)
- 9
- 10
- 11
- 12
Click Here for the Answer
Ans: (A)
The 100th and 1000th places will only have a value of 1. Unit and tens numbers can now be (1, 3), (1, 9), (3, 1), (3, 7), (5, 5), (7, 3), (7, 9), (9, 1), and (9, 7).
Ques 2: A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles. A possible value of the number of tiles along one edge of the floor is __ (CAT 2005)
- 10
- 12
- 14
- 16
Click Here for the Answer
Ans: (B)
Let m be the length and n be the breadth of the rectangle.
White tile units = W =2m+2(n−2)=2(m+n−2)
Red Tile Units = R = mn − 2 (m + n − 2)
Given that no of White tiles is equal to Red tiles.
W =R
4(m+n−2)=mn
⇒ mn−4m−4n=−8
⇒(m−4)(n−4)=8
⇒ m − 4 = 8 or 4
⇒ m = 12 or 8
Ques 3: Let n ! = 1 × 2 × 3 × ... × n for integer n ≥ 1 .If p = 1! + (2 × 2!) + (3 × 3 !) + ... + (10 × 10!), then p + 2 when divided by 11! leaves a remainder of __ (CAT 2005)
- 10
- 0
- 7
- 1
Click Here for the Answer
Ans: (D)
Let P = 1 ! = 1
Then P+2 =3 divided by 2!, 1 will be the remainder.
IF P = 1!+2 x 2! = 5
Then P+2 = 7 divided by 3!, 1 will be the remainder.
P =1!+ (2×2!)+ (3×3!)+...+ (10×10!)
When divided by 11! leaves remainder 1.
Ques 4: Let S be the set of five digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S ? (CAT 2005)
- 228
- 216
- 294
- 192
Click Here for the Answer
Ans: (B)
We would first look at the case where 1 is at the first place, 3 or 5 are at the 100th place, and an even number is at the 10000th place.
The number of possible cases is 2 x 2 x 2 = 8.
So, when 1 comes in at the unit position and an odd number comes in at the 100th position, there are 8 cases.
There are also 8 cases when an odd number comes in at the 10000th position, so the total number of times 1 comes in at the unit position is 16.
So, adding up all the cases gives us 16 (1 + 3 + 5) = 144.
Now, when an even number comes up in the first spot.
Then, 3 x 2 x 2 x 1 x 1=12 is the number of possible cases.
So, the sum of these numbers at the unit point is 12(2+4), which is 72.
At the unit position, the total of all these numbers will be 144 + 72 = 216.
Ques 5: Each family in a locality has at most two adults and no family has fewer than 3 children. Considering all the families together, there are more adults than boys, more boys than girls and more girls than families. Then, the minimum possible number of families in the locality is __ (CAT 2004)
- 4
- 5
- 2
- 3
Click Here for the Answer
Ans: (D)
Given, Number of families > Number of people > Number of boys > Number of girls > Number of households.
Let us go back to the choices and start with the one with the least value. Since the least number of families possible has been asked for.
In option (c), the number of families is 2 the number of girls 3, the number of boys 4, and the number of people 5. But the most adults that can be in two families together is 4.
∴ Count of people 2.
In option (D), the number of families is three, the number of girls is four, the number of boys is five, and the number of people is six.
There can be no more than 9 children and 6 adults in 3 groups. So, the least amount of people is 3
Ques 6: N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other. If the total time taken for singing is 28 min, what is N ? (CAT 2004)
- 5
- 7
- 9
- None of these
Click Here for the Answer
Ans: (B)
Each of the N persons from a pair with (N −3) persons (i.e. excluding the person himself and the adjacent two). So, the total number of pairs that can
be formed \(\frac{N (N-3)}{2}\)
The total time they sing = \(\frac{N (N-3)}{2}\) x 2 = 28
N(N −3)=28
N2 − 3N − 28 = 0
∴ N =7or 4
⇒ N =7(as N >0).
Ques 7: The remainder, when (1523+2323) is divided by 19 is __ (CAT 2004)
- 4
- 15
- 0
- 18
Click Here for the Answer
Ans: (C)
an + bn is always divisible by a + b [ n is odd.]
∴1523 + 2323 is always divisible by 15 + 23 = 38.
As, 38 is a multiple of 19, 1523 + 2323 is divisible by 19.
∴ We get a remainder of 0.
Ques 8: The remainder when 784 is divided by 342 is (CAT 1999)
- 0
- 1
- 49
- 341
Click Here for the Answer
Ans: (C)
Based on the remainder theorem, expression can be written in this form
\(\frac{7^{84}}{342} = \frac{(7^3)^{28}}{342} = \frac{343^ {28}}{342}\)
Remainder will be 1.
Ques 9: What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7 ? (CAT 2003)
- 666
- 676
- 683
- 777
Click Here for the Answer
Ans: (B)
Identifying the 2 digit numbers which are 10,17,24,….
Sum = 10+17+24+….+94 will be an AP
a=10, n= 13 and d=7
Sn= \(\frac{13}{2}\) [2 x 10+ (13 – 1) x 7] = 676
Ques 10: After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number? (CAT 2002)
- 80
- 75
- 41
- 53
Click Here for the Answer
Ans: (D)
The number = 3{4(7x + 4) + 1} + 2 = 84x + 53
Hence, if the number is divided by 84, the remainder is 53.
Ques 11: When 2256 is divided by 17, the remainder would be __ (CAT 2002)
- 1
- 16
- 14
- None of these
Click Here for the Answer
Ans: (A)
2256 = (24 )64 = (17 − 1)64.
In the expansion of (17 − 4)64 every term is divisible by 17 except (−1)64 which is 1.
Ques 12: Number S is obtained by squaring the sum of digits of a two digit number D. If difference between S and D is 27, then the two digit number D is __ (CAT 2002)
- 24
- 54
- 34
- 45
Click Here for the Answer
Ans: (B)
Using trial and error with the options provided in the question:-
(5+4)2 -54 = 27
So, b is the right answer.
Ques 13: A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35. What was the number erased? (CAT 2001)
- 7
- 8
- 9
- None of these
Click Here for the Answer
Ans: (A)
Let the last number of the series be n and the number erased by x
Using the formulae of series:-
\(\frac{\frac{n(n+1)}{2}- x}{n-1} = \frac{602}{17}\)
\(\frac{n (n-1)-2x}{2(n-1)}= \frac{602}{17}\)
Putting the values in the options.
Let x =7 n = 69 which is an integer.
Ques 14: The integers 34041 and 32506, when divided by a three-digit integer n, leave the same remainder. What is the value of n ? (CAT 2000)
- 289
- 367
- 453
- 307
Click Here for the Answer
Ans: (D)
Let x stand for the usual residue.
(34041- x) and (32506 x) would be completely divisible by n.
So, the difference between the two numbers, (34041 – x) and (32506 – x), will also be divisible by n,
i.e., (34041 - x - 32506 + x) = 1535.
Using the choices we have, we can see that 1535 is divisible by 307.
Ques 15: N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12 ? (CAT 2000)
- 0
- 9
- 3
- 6
Click Here for the Answer
Ans: (C)
Based on the remainder theorem
\(\frac{1421 \times 1423 \times 1425}{12}= \frac{5 \times7 \times 9}{12}= \frac{315}{12} \)
Remainder will be 3.
How to approach Remainder and Factorial questions in CAT
- Remembering all the formulae is the first step and revising every week will provide you an edge over other students.
- Remainder and Factorials are one of the easiest sections of CAT. Do not leave these questions in any exam
Practice as much questions on these concepts be
Comments