Classical

Numbers & Algebra - Numbers and Algebra MCQ

1:  

If x - y = 1, then x3 - y3 - 3xy equals

A.

0

B.

1

C.

2

D.

x2 - y2

 
 

Option: B

Explanation :

x3 - y3- 3xy = x3 - y3 - 3xy(x y)....as (x - y) = 1 
= (x - y)3= 1.


2:  

If (a, n)! is defined as product of n consecutive numbers starting from a, where a and n are both natural numbers, and if H is the HCF of (a, n)! and n!, then what can be said about H?

A.

h = a!

B.

h = n!

C.

h ≥ n!

D.

h ≥ a * n

 
 

Option: B

Explanation :

(a. n)! = product of n consecutive natural numbers starting from 'a' which is atleast divisible by n!. (n)! = product of n consecutive natural numbers. For n = 2 : (a. n)! = a(a + 1) and n! = 2 a(a + 1) is divisible by 2!. For n = 3 : (a n)! = a(a + 1)(a + 2) and n! = 6. One of the factors of a(a + 1)(a + 2) is divisible by 3 and other by 2. Thus, proceeding in this manner, (a. n)! and n! have HCF = n!  ∴ H = n!.


3:  

If a and b are prime numbers, which of the following is true?
I. a2 has three positive integer factors.
II. ab has four positive integer factors.
III. a3 has four positive integer factors.
Codes

A.

I and II only

B.

II and III only

C.

All of these

D.

None of these

 
 

Option: C

Explanation :

Factors of a2 are 1. a and a2.
Factors of ab are 1, a, b and ab.
Factors of a3 are 1. a. a2 and a3.


4:  

If x = b + c, y = c a, z = a b, then
x2 + y2 + z2 - 2xy - 2xz + 2yz is equal to

A.

a + b + c

B.

4b2

C.

abc

D.

a2 + b2

 
 

Option: B

Explanation :

x2 + y2 + z2-2xy - 2xz + 2yz 
= (x - y - z)2
=(b+c-c+a-a+b)2= 4b2


5:  

An Egyptian fraction has a numerator equal to 1, and its denominator is a positive integer. What is the maximum number of different Egyptian fraction such that their sum is equal to 1, and their denominators are equal to 10 or less?

A.

3

B.

5

C.

7

D.

9

 
 

Option: A

Explanation :

We ignore 1/7 and 1/9 because no sum of other denominator.numbers is going to give 7ths or 9ths in the denominator
Also, 1/5 and 1/10 are not enough to add up to anything (1/10, 2/10 and 3/10 are going to leave tenths left over no matter what else you add)
What's left is 1/2, 1/3, 1/4, 1/6, 1/8.
Sum total of these is 11/8. So we need all of them except 3/8, which means 1/2+1/3+1/6.
Which is the only way to do this with egyptian fractions whose denominators are 10 or less.
Hence maximum number of Egyptian fractions needed is 3




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