TCS NQT Number System Questions Quiz 1

If you need of remainder 35, while dividing a number by 460. The Unit place of that number must be 5.
lets go with options so when adding 797 so we get unit place as 5So 5678+797 = 6475
\frac{6475}{460} gives reminder 35



Video solution:

Let the number be denoted by A.

According to the given information: A, when divided by 5, gives a remainder of 0. A, when divided by 3, gives a remainder of 2. A, when divided by 2, gives a remainder of 1.

We can use this information to find the remainder when A is successively divided by 2, 3, and 5.

Let's start with the first condition: A divided by 5 gives a remainder of 0. This implies that A is a multiple of 5.

Next, when A is divided by 3, it gives a remainder of 2. This means A leaves a remainder of 2 when divided by 3.

Now, let's consider the division by 2: Since A leaves a remainder of 1 when divided by 2, we can conclude that A is an odd number.

Now, based on the above conditions, we can determine the possible remainders when A is successively divided by 2, 3, and 5:

When an odd number is divided by 2, the remainder is always 1. When a number leaves a remainder of 2 when divided by 3, the possible remainders when the number is divided by 2 can be either 0 or 1 (as 2 divided by 3 gives a remainder of 2). When a multiple of 5 is divided by any number, the remainder is always 0.

Therefore, the possible remainders when A is successively divided by 2, 3, and 5 can be 1, 0, and 0, respectively.

To summarize, when A is divided successively by 2, 3, and 5 in that order, the remainders will be 1, 0, and 0, respectively.



Video solution:

Number = LCM(12 , 27 , 35) + 6

= 3780 + 6

= 3786



Video solution:

The digital sum of number is
(1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9)=90=(9+0)=9

• So it is divisible by 3 and 9 both


• The given divisor 6561 = 9⁴. So,


• \frac{9^{24}}{9^{4}} Completely divisible so Reminder is zero

There is a standard way of solving these questions quickly, which you should mug up/write in notes:

dividend = divisor × quotient + remainder

• The remainder when (m + n) is divided by 12 is 8


• What will be the minimum possible value of (m+n) when the quotient is 1

• (m + n) = 12 x 1 + 8


• (m + n) = 20

• (m - n) = 12 x 1 + 6


• (m - n) = 18

• 2m = 38


• m = 19


• replacing m in any of the above equation we will get n = 1

This, a textbook question and you must make notes of this as its as in many exams.

• PreRequisite: How to Calculate LCM & HCF & Prime Factors

• The Prime Factors for 2088 is (2 x 2) x 2 x (3 x 3) x 29


• Here we have 1 pair of 2, 1 pair of 3


• However, one single orphan 2 and one single orphan 29

• 2088 must be divided by 2 x 29 = 58 for Perfect square

Here $ represent the - operator (subtraction)

• so X - X = 0 first condition


• X $ (Y $ Z) = X - (Y - Z) = X - Y+ Z
  
  
  
  • It can be written as X $ Y + Z
  
  
  • Example: 2012 - 0 + 2012 - 1912 = 2112
  
  
  
  

Given A = x³y² , B = xy³

A = x \times x \times x \times y \times y
B = x \times y \times y \times y

The highest common factor between A and B is xy²



Video solution:

X × Y × Z = 1800 = 2^{3} \times 3^{2} \times 5^{2} = 1800

Let

• X = 2^a1 × 3^b1 × 5^c1


• Y = 2^a2× 3^b2 × 5^c2


• Z = 2^a3 × 3^b3 × 5^c3

• Now, according the formula mentioned on this page here


• Property 4 and Formula 5


• The number of ways k identical Balls can be put inside n Distinct boxes would be: ^(k+n-1)C_(n-1)

1. So a1 + a2 + a3 = 3 which gives ^(3+3-1)C_{(3-1) =} ⁵C₂ = 10 solutions
   
   
   
   1. a1, a2, a3 as boxed i.e. n : 3
   
   
   2. K values as balls, i.e. required power: 3 i.e. K = 3
   
   
   
   


2. So b1 + b2 + b3 = 2 which gives ^(2+3-1)C_{(3-1) =}⁴C₂ = 6 solutions
   
   
   
   1. b1, b2, b3 as boxed i.e. n : 3
   
   
   2. K values as balls, i.e. required power: 2 i.e. K = 2
   
   
   
   


3. So c1 + c2 + c3 = 2 which gives ^(2+3-1)C_{(3-1) =} ⁴C₂ = 6 solutions
   
   
   
   1. c1, c2, c3 as boxed i.e. n : 3
   
   
   2. K values as balls, i.e. required power: 2 i.e. K = 2
   
   
   
   

HCF of the 148-4, 246-6 and 623-11 i.e HCF of 144, 240, 612 is the greatest number that leave remainders 4,6 and 11 respectively.
144 = 2 x 2 x 2 x 2 x 3;
240 = 2 x 2 x 2 x 2 x 3 x 5;
612 = 2 x 2 x 3 x 3 x 17
i.e HCF is 12



Video solution: