Basics of Counting & Summation  Kamal Lohia

How many 1 digit natural numbers are there?
9 (1 9)How many 2 digit natural number are there?
90 (10  99)How many 3 digit natural numbers are there?
900 (100  999)How many ndigit natural numbers are there?
I hope you got the pattern by now. 9 × 10^{n1}Palindromes are the numbers which read same forward and backward.
How many palindrome numbers are there less than 1 million?
1  Digit palindromes (a) = 9
2  Digit palindromes (aa) = 9 × 1 = 9
3  Digit palindromes (aba) = 9 × 10 × 1 = 90
4  Digit palindromes (abba) = 9 × 10 × 1 × 1 = 90
5  Digit palindrome (abcba) = 9 × 10 × 10 × 1 × 1 = 900
6  Digit palindromes (abccba) = 9 × 10 × 10 × 1 × 1 × 1 = 900
So total number of palindromes less than 1 million is = 2(9 + 90 + 900) = 2(999) = 1998.
How many times you write digit 1 while writing all natural numbers from 1 to 1000?
No need of hectic work. Just see that if we write numbers from 000 to 999, then each digit from 0 to 9 is being used at each place equally. And total number of digits being used to write these 1000 3digit numbers (from 000 to 999) is 3000. So digit 1 has been used 3000/10 = 300 times from 1 to 999 and once in 1000. So final answer becomes 300 + 1 = 301. Please read it patiently. I am sure it'll be clear to you easily.
How many times you write digit 0 while writing all natural numbers from 1001 to 2000?
This should not be any problem anymore. Answer is 300.
Just see that from 000 to 999, we write 1000 numbers using each of the digit from 0 to 9 being used at every place equally i.e. 3 × 1000/10 = 300 times each. And in this question, we have numbers from 1001 to 2000. Just remove the thousandth digit and arrange remaining digits accordingly. Isn't it simple? :)
If we write all natural numbers from 1 to 99, then how many of them contain digit 1?
Read carefully. Answer is 19 not 20 :)
While writing all natural numbers from 100 to 199, how many contain digit 1?
That was easy. It's 100 :)
Now a CAT like problem
To write all the page numbers of a book, exactly 136 times digit 1 has been used. Find the number of pages in the book.
From 1 to 99, 20 times digit 1 is used.
From 100 to 199, 120 times digit 1 is used.
So, total pages should be 4 less than 199 i.e.195.
Another (Last for now) type of counting.
All the page numbers of a book has been added and sum was found to be 1000. But teacher told that one page number has been mistakenly added 3 times. Can you identify the mistakenly added page number?
a. 5 b. 10 c. 27 d. cannot be determined
Number of pages, n should be such that n (n + 1)/2 < 1000. We get approximately n < 45.
For n = 44, Sum = 990 i.e. 10 = twice the sum of mistakenly added number (say X ) i.e. X = 5.
For n = 43, Sum = 946 i.e. 54 = 2X or X = 27.
No other values satisfy. So answer can be 5 or 27. (d)
Finding the Sum
Sum of first 'n' consecutive natural numbers is a three digit number whose digits are same. Find 'n'.
1 + 2 + 3 + ...+ n = n (n + 1)/2 = aaa = a (111) = a × 3 × 37
i.e. n (n + 1) = a × 2 × 3 × 37
As ‘a’ is a single digit number, only a = 6 satisfies. That is n = 36.
Sum of five consecutive integers is A. Find the sum of next five consecutive integers.
Each corresponding term in the next five consecutive integers is at a gap of 5 from its counterpart in previous five consecutive integers.
That means the required sum is A + 25.
In General: If sum of n consecutive integers is A, then sum of next n consecutive integers is A + n^{2}.
Sum of four 2digit consecutive odd integers when divided by 10 results in a perfect square. How many such sets of four 2digit numbers are possible?
Four consecutive 2digit numbers will have average as an even number situated between two central odd numbers. So certainly it'll be a 2digit number.
That means sum of four numbers should be of the form S = 4 × Avg but this need s to be of the form 10 × a^{2 }also.
That means Avg should be a two digit number of the form 10 × b^{2 }where b is a single digit number. (Think why??)
Only possible values for b are 1, 2 or 3 so that Avg remains a 2digit number.
But b = 1 is negated as in that case all four numbers will not be 2digit numbers.
So TWO cases are possible(i) Avg = 10 × 2^{2 }= 40 i.e. numbers are 37, 39, 41, 43 And (ii) Avg = 10 × 3^{2 }= 90 i.e. numbers are 87, 89, 91 and 93.
Sum of first n consecutive natural numbers is a perfect square for n = 1, 8. Find next two such values.
I am leaving its explanation for you to do on your own. Answer is 49 and 288.
S = (n) (n+1)/2, so either n, or n+1 has to be the 2nd multiple of a perfect square Check for all odd perfect squares and its neighbors.