All about numbers : Concepts and Results  Anubhav Sehgal, NMIMS Mumbai

First of all, here’s an overview of all types of number that would concern us in our discussions
The whole numbers mentioned here are sometimes termed as nonnegative integers and used quite frequently these days in question terminology. Now that you have an overview of existing types, let’s dive into some of these types explicitly and discover what special properties some numbers exhibit. First we will look at natural numbers and its extensions that form the base for problem content in MBA examinations.
Natural numbers and Counting Principles
 There are 9 single digit, 90 two digit, 900 three digit and so on up to 9 * 10^(n  1) ndigit numbers.
 For any consecutive 100 natural numbers, any digit 19 appears 10 times each at unit's and ten's place.[Count manually for hundredth's place]
 Sum of first N natural numbers = N*(N + 1)/2
Note a pattern here:
Sum of 110 = 55
Sum of 1120 = 155
Sum of 2130 = 255 and so on. .  Sum of first N squares = N*(N + 1)*(2N + 1)/6
Note a pattern here:
Sum of squares of 110: 385 [+2100] gives
Sum of squares of 1120: 2485 [+4100] gives
Sum of squares of 2130: 6585 [+6100] gives
...and so on.  For a range A to B,
Total numbers in the range A < = N < = B is equal to (B  A) + 1
Total numbers in the range A < = N < B OR A < N < = B is equal to (B  A)
Total numbers in the range A < N < B is equal to (B  A)  1
Let us look at some solved examples to learn how to make use of these concepts in problem solving.
Solved example 1: How many types would a typewriter press the keyboard keys to type all the natural numbers from 8 to 160?
Solution: What we are basically asked here is the number of digits that will be used for numbers 8 to 160. Single digit numbers: 8, 9. So, 2 * 1 = 2 digits or keyboard presses. Two digit numbers: We know there are 90 two digit numbers. So 90 * 2 = 180 digits or keyboard presses Three digit numbers: We have three digit numbers from 100160 to be typed. So, (160 – 100) + 1 = 61 three digit numbers to be typed. Hence, 61 * 3 = 183 digits or keyboard presses. So the keyboard keys will need to pressed a total of 2 + 180 + 183 = 365 times.
Solved example 2: A reader was reading a novel of N pages. While reading he dropped water on one of the leaves of the book and couldn’t read the page numbers on the pages spoiled. Summation of rest of the page numbers is 1010. Help him find the page numbers of the spoiled pages
Solution: We know that 110 = 55, 1120 = 155, 2130 = 255, 3140 = 355. Adding them up we have: 140 = 820. Next is 4150 = 455. But that will cross our summation by a long margin. Hence we count from here manually until the sum just crosses our sum. This is done as the difference between the ‘would have been’ sum with no spoiled pages and the given sum will provide us the sum of consecutive page numbers that are spoiled.
820 + 41 + 42 + 43 + 44 = 990 990 + 45 = 1035 1035 – 1010 = 25. The missing page numbers will be 12, 13.Solved example 3: How many times will the digit 4 be used in writing all natural numbers from 1 to 450.
Solution: We know that 4 appears 10 times at unit’s digit and 10 times at ten’s digit for any 100 consecutive natural numbers.
So we have: 1  100: 4 appears 20 times
101  200: 4 appears 20 times
201  300: 4 appears 20 times
301  400: 4 appears 20 times [We are not considering the hundredth’s place right now]
401  450: 4 appears 15 times [We are not considering the hundredth’s place right now]
Also, 4 appears for (450 – 400) + 1 = 51 times at hundredth’s place
So digit 4 will be used a total of 20 * 4 + 15 + 51 = 146 times.Next we look at properties and concepts related to perfect squares and prime numbers in general that form the base concepts for intriguing problems of this domain.
Perfect Squares
 A perfect square cannot end with 2,3,7,8 or odd number of zeroes.
 Every perfect square has even ten's digit except the ones which end with 6.
 Perfect squares have only 22 different possibilities for last two digits which are deduced from squares 125.
 Digital sum of a perfect square can only be 1, 4, 7, or 9.
 Patterns:
i) Square with largest occurrence of a nonzero digit at end is 38^2 = 1444 with 4 coming thrice.
ii) Only square of the form aa^2 = bbcc is 88^2 = 7744
iii) 1^2 = 1; 11^2 = 121; 111^2 = 12321; 1111^2 = 1234321; and so on..
iv) 1 and 9 are the only perfect squares with all the digits odd.  For any nonnegative value of x ,
x^2 > x for x > 1
x^2 < x for x < 1 [NOTE : non negative hence range is 0< x < 1]
x^2 = x for x = 0,1  A perfect square is either of 3k or 3k + 1 form ; 4k or 4k + 1 form and similarly gives only a few select remainders for other divisors as well.
How to check in general:
Check for 3 :
1^2 = 1 ; 1 mod 3 = 1
2^2 = 4 ; 4 mod 3 = 1
3^2 = 9 ; 9 mod 3 = 0
Check only for the cycle of each number (so upto 3^2 for 3,upto 4^2 for 4..and so on.)
Prime Numbers
 There are infinite prime numbers
 There are infinite number of twin primes (The set of N, N+2 such that both of them are primes)
 There is exactly one prime triplet N, N+2, N+4 where N = 3.
 A prime number p, p > 3, is always of the form 6k +/ 1 and is not divisible by any prime M such M < = √ p
 110: 4 primes; 1100: 25 primes; 110^3: 168 primes; 1 10^4: 1229 primes [Just an extra information. Sometimes useful.]
 If A, B are primes then either A  B or A + B is divisible by 3.
Also, if A, B, A  B, A + B are all primes then only one solution is there i.e. A = 5, B = 2.  3
31
331
3331
33331
333331
3333331
33333331 are all primes while 333333331 is not