CAT Questions On Factorial PDF Set-2

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CAT Questions On Factorial PDF Set-2
CAT Questions On Factorial PDF Set-2

CAT Questions On Factorial PDF Set-2:

Download CAT Questions On Factorial PDF Set-2. Practice important Factorial Questions with detailed answers and explanations. These questions are based on previous CAT question papers.

Download CAT Questions On Factorial PDF Set-2

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Question 1: A 80 digit number has all 7’s in it. What is the remainder when this number is divided by 74?

a) 7
b) 37
c) 3
d) None of these

Question 2: How many natural number values of x exist such that $\frac{(2x+1)^2}{2x+7}$ is an integer.

a) 1
b) 2
c) 3
d) 4

Question 3: What is the remainder when $(26)^7+(27)^7+(28)^7+(29)^7$ is divided by 110?

a) 1
b) 2
c) 4
d) 0

Question 4: Question 4: What is the number of ways in which 2016 can be divided into two factors such that at least one of them is less than 40?

a) 16
b) 17
c) 20
d) 25

Question 5: If A=7!+ 8!+ 9!+ 10!+ 11!, which of the following is a factor of A?

a) 1001
b) 897
c) 1771
d) 2261

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Answers & Solutions:

1) Answer (C)

77 mod 74 = 3, 777 mod 74 = 37, 7777 mod 74 = 7.
After this step the remainder starts repeating itself.
777…80 times mod 74 = 3

2) Answer (B)

$\frac{(2x+1)^2}{2x+7} = \frac{(2x+7)(2x-5)+36}{2x+7} = 2x-5 +\frac{36}{2x+7}.$
2x+7 should be the multiple of 36.
only x = 1 satisfy this condition.

3) Answer (D)

When $n$ is odd, $a+b$ is a factor of $a^n+b^n$

So, $26^7+29^7$ is divisible by $26+29=55$
Similarly $27^7+28^7$ is divisible by $27+28=55$

Note that $26^7+29^7$ and $27^7+28^7$ are both odd and their sum is hence even.

Therefore, $(26)^7+(27)^7+(28)^7+(29)^7$ is even and is divisible by 55. Therefore, it is divisible by 55*2=110

Hence the remainder would be 0.

4) Answer (B) 

In order to solve this problem, let us factorize 2016 first.
$2016 = 2^5*3^2*7$
The number of ways this can be factored into two numbers is $\frac{1}{2}\times (5+1)\times(2+1)\times(7+1) = \frac{1}{2}\times 6 \times 3 \times 2 = 18$
Note that $\sqrt{2016} \approx 45$. So, most of the factorizations have at least one number less than 40.
The only exception to this is when 2016 is factorized as 48*42.
Hence, the number of ways of factorizing 2016 such that at least one of the factors is less than 40 is 18-1 = 17

5) Answer (D)

A=7!+ 8!+ 9!+ 10!+ 11!
=> A=7!(1+ 8+ 8X9+ 8X9X10+ 9X10X11)
=> A= 7 X 720 X (1+8+72+720+7920)
=> $7 \times 720 \times 8721$
=> $3^{3} \times 720\times 7 \times 19\times17$
=> $3^{3}\times 720 \times 2261$

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