# Numbers and digits - Atreya Roy

• Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal.

Concept : When questions of numbers (2 digit or 3 digit) are given and there are a few set of Restrictions, we will  take the unit digit to be “b”, tens digit to be “a”, Hundreds digit to be “c” and so on.

So for a 2 digit number, It may be represented as : 10a + b

For a 3 digit number : 100c+10a+b

And So on….

Example:

How many 2 digit possible numbers are possible if the digits when interchanged, The absolute  difference between the two 2-digit numbers is 63 ?

Solution :

Let the number be N. N = 10a+b

Reversing N we get N’= 10b+a

Difference between them = 10a+b – (10b+a)= 9(a-b)

As said earlier, difference between the two numbers = 63

Hence 9(a-b) = 63

a-b = 7

Possibilities:

a=b=0 not possible, since both of them are said to be 2 digit.

A= 8, b=1

A=9, b=2

And their reverse numbers as well

Hence 4 such numbers possible.

Finding the unit digit

The Concept of unit digit of a number is as important as any theory of Permutation Combination. The application of unit digit is enormous. We can use the concept of unit digit solve sums in seconds that would have taken us minutes working in the traditional approaches. So today we will go through all these concepts.

Cyclicity of 2 : 2^1 = 2 , 2^2 = 4, 2^3 = 8, 2^4 =16 , 2^5 = 32

So we can see that for every 4 set of exponents starting from 1, the unit digit of 2^n remains same. Hence We will call it as cycle order =4 (After every 4 terms, the unit digit is repeated)

Example : What is the unit digit of 2^73

Solution : Find 73 mod 4 (Because cyclicity of 2 = 4) 73mod 4 =1.
Hence 2^73 ‘s unit digit will be same as 2^1 = 2

Cyclicity of 3 : 3^1 =3; 3^2 = 9; 3^3 = 27 ; 3^4 = 81 ; 3^5= 243.
Hence again, after 4 terms starting from 1, the unit digit of powers of 3 get repeated. Hence the cycle order for 3 = 4

Cyclicity of 4 : 4^ odd always ends with 4 / 4^even always ends with 6 = > Hence the cycle order of 4 = 2

For 5 and 6, 5^a and 6^b : where a and b are positive integers, both will end in 5 and 6 respectively.
Similarly we can find the cyclicity of other numbers

Example :

What is the unit digit of the sum : 5^95+ 6^39

Solution: Last digit of 5^95 = 5

Last digit of 6^39 =6

Unit digit of the sum = 5+6 = 1 1

Hence the Unit digit =1

Note : Always consider the last digit of the number, not the whole number. Calculations become easy. But the exponent remains same.

Example: Find the unit digit of 26*31*48*37*94

Solution :

Consider the last digits: 6*1*8*7*4 = 6*8*7*4 = 4 8 *7*4 = 8*7*4 = 6*4 = 4

Example: Find the unit digit 5^49*2^4*7^48*9^34

Solution :

5^49 will give 5 as its unit digit. 2^4 =16

So, 5*even = 0

So the unit digit = 0

Examples : What is the unit digit of 1! +2! +3!.....100! ?

Solution: Numbers after 4! : 5!,6!.. till 100! Will have 0 as their unit digit.

So the unit digit of the series =unit digit of the sum till 4! = 1+2+6+24 = 33

Hence of the unit digit = 3

Finding number of digits in a^b

We might often see questions about the total number of digits in a number which is denoted as a^b. Doing these manually is quite impossible and we don’t want to give much time to solve it. Hence we will today learn how to find the number of digits using the method of Logarithm.

How to find the Number of digits in a^b ?

Number of digits = |_ 1+ log (a^b) _|

Here |_ x _| denotes the floor function

We will calculate the value of log(a^b) in base 10

Example :

What is the number of digits in 35^29

Solution: |_ 1+ log 10 (35^29) _|

= |_ 1+ 29 log 10 (35)  _|

= |_ 1+ 29*1.54 _|

= |_ 1+ 44.776 _|

= |_ 45.776 _|

= 45 digits

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