# Types of Questions

Questions from this concept are asked in three different ways:- (Base)
_{10}to any other base and vice-versa. - (Base)
to (Base)_{x}and vice-versa; none of_{y}*x*and*y*being equal to 10 but*x*and*y*will be given. - (Base)
_{x}, value of_{y}*x*and*y*will not be given.

# (Base)_{10} to any other base and vice-versa

**Method 1**

Let us see in case of (74)

_{10}.(74)

_{10}= 7 Ã— 10^{1}+ 4 Ã— 10^{0}, since the base is 10.Now, if we have to convert this number in 9 base, then we will try to write it in terms of powers of 9.

(74)

_{10}= 8 Ã— 9^{1}+ 2 Ã— 9^{0}= (82)_{9}(74)

_{10}= 1 Ã— 8^{2}+ 1 Ã— 8^{1}+ 2 Ã— 8^{0}= (112)_{8}(74)

_{10}= 1 Ã— 7^{2}+ 3 Ã— 7^{1}+ 4 Ã— 7^{0 }= (134)_{7}(74)

_{10}= 2 Ã— 6^{2}+ 0 Ã— 6^{1}+ 2 Ã— 6^{0}= (202)_{6}While converting the numbers from decimal system to any other system of writing the numbers, we should be concerned with the following two rules

- Take maximum possible power of the base and then keep writing rest of the number with the help of lesser power of base (as illustrated in the above example).
- Once we have used (base)
; where n is the maximum power, we will be required to^{n}

E.g., Write the co-efficients of all the powers of base from 0 to (
Now, suppose we have to convert (356)
(356)

*n*âˆ’ 1) as in the case of (74)_{10}^{ }= (202)_{6}._{7}in the base of 10._{7}= 3 Ã— 7^{2}+ 5 Ã— 7^{1}+ 6 Ã— 7^{0}= (188)_{10}**Method 2**

Converting (74)

_{10}to the base of ( )_{9}_{}

So, (74)

Converting (74)

_{10}= (82)_{9}_{10}to the base of ( )_{8}_{}

So, (74)

_{10}= (112)_{8}_{}Converting (74)_{10}to the base of ( )_{7}_{}

So, (74)

_{10}= (134)_{7}_{}Converting (74)_{10}to the base of ( )_{6}_{}

So, (74)

_{10}= (202)_{6}

# (Base)_{x} to (Base)_{y} and vice-versa; none of x and y being equal to 10 but x and y will be given

Converting (345)_{8}to the base of ( )

_{9}

_{}

We will do this problem with the help of creating a bridge of base 10 between base 8 and base 9.

345 = 3 Ã— 8

(229)

However, if new base is a power of old base and vice-versa, then it can be converted directly also in the new base, i.e., we are not needed to go to base 10 for these kind of conversions.
E.g., (Base)
Converting (101110010)

At first, we will club three digits of binary number into a single block and then write the decimal equivalent of each group (left to right).

So, (101110010)

Now, (101)

(110)

(010)

So, (101110010)

At first, we will club four digits of binary number into a single block and then write the decimal equivalent of each group (left to right).
So, (101110010)

Now, decimal equivalent of (0001)

Decimal equivalent of (0111)

Decimal equivalent of (0010)

(101110010)

**Convert (345)***Step 1*_{8}into base 10.345 = 3 Ã— 8

^{2}+ 4 Ã— 8^{1}+ 5 Ã— 8^{0}= (229)_{10}*Step 2***Now convert this number in base 10 into base 9.**(229)

_{10}= 2 Ã— 9^{2}+ 7 Ã— 9^{1}+ 4 Ã— 9^{0}= (274)_{9}However, if new base is a power of old base and vice-versa, then it can be converted directly also in the new base, i.e., we are not needed to go to base 10 for these kind of conversions.

_{2}to (Base)_{4}or (Base)_{2}to (Base)_{8}conversion does not require a bridge of base 10._{2}to Octal ( )_{8}systemAt first, we will club three digits of binary number into a single block and then write the decimal equivalent of each group (left to right).

So, (101110010)

_{2 }is now (101)_{2}(110)_{2}(010)_{2}Now, (101)

_{2 }= 1 Ã— 2^{2}+ 0 + 1 Ã— 2^{0}= 5(110)

_{2}= 1 Ã— 2^{2}+ 1 Ã— 2^{1}+ 0 Ã— 2^{0}= 6(010)

_{2}= 0 Ã— 2^{2}+ 1 Ã— 2^{1}+ 0 Ã— 2^{0}= 2So, (101110010)

_{2 }= (562)_{8}_{}Converting (101110010)_{2}to Hexa-decimal ( )_{16}systemAt first, we will club four digits of binary number into a single block and then write the decimal equivalent of each group (left to right).

_{2 }is now (0001)_{2 }(0111)_{2 }(0010)_{2}Now, decimal equivalent of (0001)

_{2 }= 1Decimal equivalent of (0111)

_{2}= 7Decimal equivalent of (0010)

_{2}= 2(101110010)

_{2 }= (172)_{16}

# (Base)_{x} to (Base)_{y} , value of x and y will not be given

In these type of questions, normally some calculation is given in some unknown system of writing numbers and on the basis of that we will be required to solve the questions.Example

In a system of writing of N digits, 4 Ã— 6 = 30 and 5 Ã— 6 = 36. What will be the value of N = 3 Ã— 4 Ã— 5 in the same system of writing?

Solution

Let us assume that there are
So, (30)

â‡’ 3

â‡’
So, this system of writing has 8 digits.

In this system 3 Ã— 4 Ã— 5 = 60 will be written as 74. (60 = 7 Ã— 81 + 4 Ã— 80)

Alternatively, since this system is having 6 as one of its digits, minimum value of

*N*digits in this system of writing.*N*= 3 Ã—*N*^{1}+ 0 Ã—*N*^{0}= 24â‡’ 3

*N*= 24â‡’

*N*= 8In this system 3 Ã— 4 Ã— 5 = 60 will be written as 74. (60 = 7 Ã— 81 + 4 Ã— 80)

Alternatively, since this system is having 6 as one of its digits, minimum value of

*N*will be 7. Again, 24 is written as 30 in this system, so*N*is less than 10. Now use hit and trial for*N*= 7 or 8 or 9 to find out N in 24 = (30)_{N}.