Important Sequences and series and their sums

Below are posted some important results which over the years I've found to be very helpful in solving problems or simply exploring problems in math.

Basic results:

Combinatorial sums:

Trigonometric series sums:

Infinite series results:

Most of these series are commonly used in math, therefore these can be found in any standard math book.

Divisibility Revised

Every natural number (greater than 1) is divisible by two natural numbers, i.e. 1 and itself. Therefore, every natural number greater than 1 has at least two factors. Prime numbers have only two factors.

Note: 1 is neither a prime nor a composite number.

Divisibility hints:

2- Any natural number is divisible by 2 when the the digit at its unit place is any one among 0,2,4,6,8. Example: 22,4,36.

3- A natural number is divisible by 3 when the sum of its digits is divisible by 3.
For example, To check if 35892 is divisible by 3, we find the sum of digits.
3+5+8+9+2=27
27 is divisible by 3. Therefore, 35892 is divisible by 3.

4- A natural number is divisible by 4 when the number formed by digits at ten's and unit's place respectively is divisible by 4.
For example, To check if 35892 is divisible by 4:
The number formed by ten's and unit's place is 92
92 is divisible by 4. Therefore, 35892 is divisible by 4.

5- Any natural number is divisible by 5 when the the digit at its unit place is any one among 0 or 5. Example: 25, 93670.

7- A natural number is divisible by 7 when the difference between twice the digit at unit's place and the number formed by rest of the digits is 0 or a multiple of 7.
For example, Consider 196.
Twice the digit at unit's place=2*6=12
Number formed by rest of the digits=19
19-12=7
Therefore, 196 is divisible by 7.

11- Take the sum S1 of the digits at odd places and the sum S2 of the digits at even places. If the difference between S1 and S2 is 0 or any multiple of 11, the number is divisible by 11.
For example, To check if 3852101 is divisible by 11:
S1= 1+1+5+3=10, S2= 0+2+8=10, S1-S2=0
Therefore, the given number 3852101 is divisible by 11.

2^n- If the number formed by last n digits of the natural number is divisible by 2^n, the number is divisible by 2^n.
For example,
If the number formed by last 4 digits of a natural number is divisible by 16, the natural number is divisible by 16. Consider 7856981600.
Similarly, If the number formed by last 5 digits of a natural number is divisible by 32, the natural number is divisible by 32.

5^n- If the number formed by last n digits of the natural number is divisible by 5^n, the number is divisible by 5^n.
For example,
If the number formed by last 2 digits of a natural number is divisible by 25, the natural number is divisible by 25. Consider 625.

3^n- If the sum of the digits of a natural number is divisible by 3^n, the natural number is divisible by 3^n.
For example,
If the sum of the digits of a natural number is divisible by 9, the natural number is divisible by 9. Consider 728109.

6^n- If a natural number is divisible by 2^n and 3^n, it is divisible by 6^n.
For example,
If a natural number is divisible by 4 and 9, it is divisible by 36.

10^n- If a natural number has n consecutive zeroes beginning from the unit's place of the number, it is divisible by 10^n.
For example,
If a number has 3 consecutive zeroes beginning from the unit's place, it is divisible by 1000. Consider 378000.

Note: If a natural number is divisible by two co-prime numbers, it is also divisible by the product of the co-prime numbers.
Thus, if a natural number is divisible by 3 and 4, it is also divisible by 12. Consider 576.
The vice versa of the statement is also true.
Thus to check if a number is divisible by 42, it suffices to check if it is divisible by 6 and 7.

- Rishabh Bidya

Ratio of the areas of the circumcircle and the incircle of an n-sided regular polygon

Question:

To find the ratio of the areas of the circumcircle and the incircle of an n-sided regular polygon.

Solution:

Result:

The ratio of the areas of the circumcircle and the incircle of an n sided regular polygon is square of the secant ratio of (pi divided by the number of sides of the regular ploygon).

Note:

The ratio of the areas of the circumcircle and the incircle of an n-sided regular polygon is independent of size of the side. It depends only on the number of sides.

© Rishabh Bidya

Difference between the areas of the circumcircle and the incircle of an n-sided regular polygon

Question:

To find the difference between the areas of the circumcircle and the incircle of an n-sided regular polygon.

Solution:

Result:

The difference between the areas of the circumcircle and the incircle of an n sided regular polygon is pi times square of the length of the side divided by 4.

Note:

The difference between the areas of the circumcircle and the incircle of an n-sided regular polygon is independent of the number of sides. It depends only on the length of the side.

© Rishabh Bidya

Pattern in fifth powers of numbers : Last two digits of fifth powers of numbers at a difference of 20

The procedure:

Take any number (Say x).

Form a set of numbers by adding 20 (Your set={x,x+20,x+40,x+60,...}).

Raise every member of the obtained set to its fifth power.

Observe the last two digits (taken from right to left) of the obtained fifth powers.

The last two digits of the fifth powers of every member of the set are the same.

The Proof:

Any term of the series will be of the format x+20k(where k must be a non negative integer).

We need to prove that the digits at the unit's and tens's place of the fifth powers of x and x+20k are same.

Example:

We choose 3 as our x.

The set of numbers obtained is {3,23,43,64,...}.

We raise the terms of this set to their fifth powers {243,6436343,147008443,992436543,...}.

We observe the last two digits of the obtained fifth powers.

The number at unit's place is 3 whereas the number at the tens' place is 4.

Derived Corollary:

The last two digits of an odd number's fifth power and another number's fifth power are the same when another number is formed by adding a multiple of 20 to the first number.

Q.E.D

© Rishabh Bidya

Magic Squares 4 : Creating a 4x4 magic square using 16 consecutive integers given the sum of a row or a column

Problem Background : A 4x4 magic square is an arrangement of 16 squares arranged in 4 rows and 4 columns. We are given a sum S which is the sum of any row or column. We need to arrange 16 consecutive integers in the square.

Solution : If the sum of a row is S, S also needs to be the sum of any other row or any other column.

Derived corollary : It can be noted that for the numbers in individual squares to be positive integers, S must be of the form 2(mod4) and S must be greater than or equal to 34.

Note that this solution is however only a special case of the general solution:

Observations:
The numbers in individual squares are positive integers when:

1. S>30d
2.  If d is even, S is divisible by 4. If d is odd, S is of the form 2(mod4)

The above result can be derived from  Magic Squares 3 :

Thus, S=4a+30d. After this, we merely substitute for a and simplify.

Q.E.D

© Rishabh Bidya

Magic Squares 3 : Creating a 4x4 magic square using 16 terms in Arithmetic Progression

Problem Background : A 4x4 magic square is an arrangement of 16 squares arranged in 4 rows and 4 columns. We are given 16 terms in an arithmetic progression(AP) to place in each of the squares such that the sum of each of the rows and columns is equal.

Solution : We assume the terms of the AP are a,a+d,a+2d,...,a+14d,a+15d where a is the first term and d is the common difference.

General Solution of a 4x4 magic square

Derived corollary : It can be noted that we get a magic square of any 16 consecutive integers when d=1 and a is the first term among the 16 integers.

Magic square of any 16 consecutive integers where a is the first integer

Example: If a=1 and d=1, we get a magic square of the integers from 1 to 16.

Magic square of first 16 natural numbers

Q.E.D

© Rishabh Bidya

Surprise your buddies: A wonderful number trick

The trick:

We tell our friend to think of any random number. Then perform some random operations on it and .. Surprise! We tell him the final answer.

Note: Before we start the trick, we must think of two numbers, a and m. a is the number that we want as our answer and m is an arbitrary multiplier. Preferably a and m should be small for our convenience during arithmetic operations. Of course, we never disclose a and m to our friend before or during the trick !!

Let us begin !!

i) Think of a and m.
ii) Multiply a and m. P=a*m.

iii) Tell your friend to think of any number, say x.
iv) Tell him to multiply x with m.
v) Tell him to subtract P from the obtained product of x and m.
vi) Now, tell your buddy to divide the obtained difference by m.
vii) Now, you tell your friend to subtract the obtained quotient from x.

TADA !! Tell your friend - The answer is a.

Let us try the trick (An example) :

i) I think of two numbers:
   a=3. (3 should be my final answer.)
   m=2.
ii) P=6.

iii) You think of a number, say 7.         (x=7)
iv) Multiply 7 with 2. Result is 14.
v) Subtract 6 from 14. Result is 8.       (P=6)
vi) Divide obtained difference by 2, i.e. 8 by 2. Result is 4      (Here our m=2. Therefore, we divide difference by 2).
vii) Subtract obtained result from the number you had thought first, i.e. Subtract 4 from 7.

The answer is 3 !!!



© Rishabh Bidya

Total number of triangles in an inverted triangle system

Question:

To find total number of triangles in an inverted triangular system which has n triangles, each of which is inverted and contained inside its previous triangle.

Example: 

Solution : 

If the total number of triangles is n, 

Except for the innermost triangle, every triangle made results in three smaller triangles, so the number of such smaller triangles is 3*(n-1)

Apart from the smaller formed triangles, there are n triangles that form the figure.

Thus, the total number of such formed triangles will be the sum of both these numbers, i.e. 3*(n-1)+n

Therefore, we obtain the total number of triangles in an inverted triangle system is 4n-3

© Rishabh Bidya

Magic Squares 2 : Creating a 3x3 magic square using 9 consecutive integers given the sum of a row or a column

Problem Background : A 3x3 magic square is an arrangement of 9 squares arranged in 3 rows and 3 columns. We are given a sum S which is the sum of any row or column. We need to arrange 9 consecutive integers in the square.

Solution : If the sum of a row is S, S also needs to be the sum of any other row or any other column.

Derived corollary : It can be noted that for the numbers in individual squares to be integers, S must be divisible by 3, i.e. S must be 0(mod3) and S must be greater than 12.

The above result can be derived from  Magic Squares 1 :

After this, we merely substitute for a and simplify.

Q.E.D

© Rishabh Bidya

Magic Squares 1 : To create a 3x3 magic square given 9 consecutive integers

Problem Background : A 3x3 magic square is an arrangement of 9 squares arranged in 3 rows and 3 columns. We are given 9 consecutive integers to place in each of the squares such that the sum of each of the rows and columns is equal.

Solution : If the given integers begin from a, rest if the integers are a+1, a+2, a+3, ..., a+7, a+8.

In the above figure, Sum of every row as also sum of every column is 3a+12.

Q.E.D

© Rishabh Bidya

Total number of triangles in a figure with h horizontal non intersecting lines and v vertical lines dropped from a vertex to its opposite side(cevians)

Question:

To find total number of triangles in a triangular body which has h horizontal lines, two of which do not intersect and v vertical lines which are all dropped from one vertex to its opposite side.

Solution:

The total number of triangles in the figure is:

(h+1)*(h+2)*(v+1)/2

Example :

If h=3 and v=2,

substituting the values of h and v in the formula given above, we get,

Total number of triangles = 30


Note : If a line is dropped from a vertex to its opposite side, it is called as a cevian.

© Rishabh Bidya

Checking if a number is prime - A faster way

In order to check if a number(say n) is prime or not, you take the following steps:

1. If n is even and n!=2, it is not a prime.
2. Check for the square root of n, say x. 
3. If x is not divisible by any prime number less than x, it is a prime.
4. Again while checking for all prime numbers less than x, you only need to look for the odd numbers. 

Example, 

1. n=9923. We check if 9923 is prime or not.
2. Since 9923 is odd, 2 is discarded.
3. Square root of 9923 is roughly 99.614, we round it off to 100.
4. Now we check if any prime less than 100 divides 9923.
5. Since we have to check for primes, for convenience, we opt for a larger subset, all odd numbers.
6. Set A={3,5,7,9,11,...,91,93,95,97,99}.
7. We check if any of these numbers divides n.
8. More efficiency is possible if we apply Sieve of Eratosthenes and strike out the numbers divisible by 3,5, and 7. We only will need to check the divisibility from the remaining numbers.

Since, we didn't check for all the numbers from 1 to 100 and only the odd ones, it made the program 200% efficient. Forget checking all numbers from 1 to 9922.

© Rishabh Bidya