Using 2,0,1,6 and arithmetic operations to get every digit from 0 to 9

2x0x1x6=0

(2^0)^16=1

2+0x1x6=2

-(2+0+1-6)=3

20-16=4

-(2x0+1-6)=5

2x0x1+6=6

2x0+1+6=7

2+0x1+6=8

2+0+1+6=9

Thus, 2,0,1,6 (or 2016 ;) ) gives us 0 to 9 (Everything in numbers). I wish 2016  to provide you 'EVERYTHING' in terms of joy, desires, goals and love (and all else that you seek). Wishing you a 'HAPPY NEW YEAR' :D Cheers!! Enjoy

© Rishabh Bidya

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 sixth powers of numbers : Last two digits of sixth powers of numbers at a difference of 10

The procedure:
Take any number (Say x).
Form a set of numbers by adding 10 (Your set={x,x+10,x+20,x+30,...}).
Raise every member of the obtained set to its sixth power.
Observe the last two digits (taken from right to left) of the obtained sixth powers.
The last two digits of the sixth powers repeat after every five members of the set.

(Note the numbers highlighted in yellow repeating in the same column)

The Proof:
Any 6 terms of the series will be of the format x,x+10,x+20,x+30,x+40,x+50.
We need to prove that the pattern of the sixth powers repeats after every 5 terms.
Therefore, it suffices to prove that the digits at the unit place and tens place of sixth power of a number are equal when the number after five terms is raised to the sixth power.
Therefore, we need to prove that the last two digits of the sixth power of x and the sixth power of x+50 are equal. (Since the numbers are taken at a difference of 10 and there are 5 such terms.)

Similarly, last two digits of sixth powers of x+10 and x+60 will be equal.
Similarly, last two digits of sixth powers of x+20 and x+70 will be equal.
And so on ...
Thus, the pattern continues to repeat after every five terms.

Derived Corollary:
The last two digits of a number's sixth power and another number's sixth power are the same when another number is formed by adding a multiple of 50 to the first number.

Important Observation:
If any number has 5 at its unit place, its sixth power ends with 2 and 5 at ten's and unit's place respectively. (This, too can be proved easily).
A similar pattern is observed in the squares, cubes, fourth powers and fifth powers of numbers.

Q.E.D

© 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