TCS Number System and HCF and LCM Quiz-1

Solution :-

Remainder in case of 148 = 4

Number = 148 - 4 = 144

Remainder in case of 246 = 6

Number = 246 - 6 = 240

Remainder in case of 623 = 11

Number = 623 - 11 = 612

Now, we have find the H.C.F. of 144. 240 and 612

_____________
2  | 144, 240, 612
|____________
2  |   72, 120, 306
|____________
3  |   36,  60,  153
|____________
|   12,  20,  51
|

H.C.F. of 144, 240 and 612 = 2*2*3 = 12

Therefore, the required number is 12 that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.

Video solution:

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*x*x*y*y
B = x*y*y*y

The highest common factor between A and B is xy²

Video solution:

X × Y × Z = 1800 = 2³ × 3² × 5²

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
   
   
   
   

1. ⁵C₂ *⁴C₂*⁴C₂ = 10*6*6 = 360

Maybe, a few of you solved the question incorrectly. The reason being you didn't read the question correctly.

The word important in the question is : Successively

• Let the number be A


• So, when divided by 5 : A = 5K + 0 = 5K


• Thus, when you divide A by 5 you get K

• From above when A is divided by 5. We get K


• So, dividing successively, K is divided further by 3 and leaves remainder 2


• So, following is valid : K = 3J + 2

• From the previous step we get J, after K is divided by 3.


• Thus, dividing successively, J is divided further by 2 and leaves remainder 1


• So, following is valid : J = 2L + 1

• J = 2L + 1


• K = 3J + 2


• A = 5K

• A = 5K


• A = 5(3J + 2) = 15J + 10


• A = 15(2L + 1) + 10 = 30L + 25

• Num : 30L + 25 (Where L can be 0, 1 , 2, 3 ......... )

• Dividing 25 by 2 we get
  
  
  
  • 12 and 1 remainder
  
  
  
  


• Dividing 12 by 3 we get (Since, successively)
  
  
  
  • 4 and 0 remainder
  
  
  
  


• Dividing 4 by 5 we get (Since, successively)
  
  
  
  • 0 and 4 remainder
  
  
  
  

Dividend = divisor × quotient + remainder
5678 = 460 ×12 + 158
Here 158 is the remainder. we need the remainder to be 35 by adding a number.
Let the number to be added be x.

Since 158>35, the quotient we will get by dividing  (5678+x) by 460 should be greater than 12. It may be 13, 14, 15 etc. But i am taking the lowest one here i.e.13.

So (5678 +x) = 460 × 13 + 35  (here 35 is remainder)
⇒ 5678 + x - 35 = 460 × 13
⇒5643 + x = 460 × 13
⇒ x = 460 × 13 - 5643
⇒ x = 337

However, none of the options match right ???

• If rather than 13th number as quotient, we would've used 14th then?


• Thus, add 460 (Divisor) to 337, we get 797

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 or 9


• The given divisor 6561 = 9⁴. So,


• (9²⁴)/(9⁴) Reminder is 0

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*1 + 8


• (m + n) = 20

• (m - n) = 12*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*2)*2*(3*3)*29


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


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

• So, 2088 must be multiplied with 2*29 i.e. 58