Test series

Question

(22)₄ + (101)₃ – (20)₅ = (x)₄ + (4)_x–1. The value of x is ___________.

Comments

@mcjoshi and @Debashish Deka

Yea and yes

@habib

I appreciate that you got my point......The value of (x)₄ is 12₄ therefore 12 is in base 4....

---

Then, bro you are wrong.( you can't use one value at one point and another value at other place) to satisfy an equation.

---

@mcjoshi

bro.... they are both the same values with different BASE/RADIX (x₄ = 12₄ = 6₁₀) for (x)₄ the value of x =12 and for 4_x-1, since x-1 is a BASE we need to denote it in BASE 10....(read the big comment above for full clarification)

Answer 1

Here we have one basic to remember : Given a base r , any digit of a number in that base cannot be greater than (r-1).

Now lets first calculate the LHS expression which gives :

(2*4 + 2) + (1*3² + 1) - (2*5) = 20 - 10 = 10

Now we are given 2 terms in RHS.Lets first focus on 2nd term.Since only digit is mentioned i.e 4 in base 'x-1' , so minimum possible value of 'x-1' has to be 5 or 'x' to be 6 using the rule which was mentioned earlier.Now that we have only 1 digit 4 in the 2nd term so its value is going to be 4 only.

Hence the value of 1st term has to be 10 - 4 = 6

But if we use 6 in 1st term it is invalid since the base mentioned is 4 hence the max possible value of any digit is 3 for 1st term.

Hence , basically what we are left with is writing the 6 which is the decimal value into its corresponding base 4 value.So

(6)₁₀ = (0110)₂ = (12)₄

Hence the value of 'x' is 12 and base of 4 which is 2nd term is 12 - 1 = 11

Comments

Ace test series gives any kind of questions.U should keep ur fundamentals clear first.

---

@habibkhan nice explaination!

Yes, Answer cannot be $6$

---

why are u writing 6 in base 4??i did nt get this point

Answer 2

from 

 RHS x < 4 and x-1>4 i.e. x>5

hence no possible value of x can satisfy this equation

Comments

Plz have a look at my explanation

Answer 3

ans in coming to be 6

Comments

6 will violate the base rule as (6)₄ is not possible