Monday, 1 June 2015

Percentage







Percentage means a rate, number or amount in each hundred.
x percentage is written as x% and expressed as x/100.

Hence, ( a/b ) is expressed as percentage = ( a/b * 100 )

Example :

20 % =  ( 20/100 ) = ( 2/10 ) = ( 1/5)

1.25 % = ( 1.25/100 ) = ( 125 / ( 100 * 100 ) ) = ( 5 / ( 4 * 100 )) = ( 1/100 ) = 0.01


Some facts:

(1) Population Calculations:

Let the population of a town is P and it increases at the rate of R % per annum then,
Population after n years = P * ( 1 + ( R/100) )ⁿ
Population n years ago = P / ( 1 + ( R/100) )ⁿ

(2) Depreciation of a product:

Let the present value of a product is P and it depreciates at the rate of R % per annum then,
Value of the product after n years =  P * ( 1 - ( R/100) )ⁿ
Value of the product n years ago = P / ( 1 - ( R/100) )ⁿ


Tuesday, 8 July 2014

Solutions: HCF & LCM, Part-A





The solutions here are posted for the questions in:
http://arithmophobia-elixir.blogspot.in/2014/07/questions-on-hcf-lcm.html


(1)  We know the formula:

HCF of Fractions = ( HCF of Numerators ) / ( LCM of Denominators )
LCM of Fractions = ( LCM of Numerators ) / ( HCF of Denominators )

Given Fractions:  2/3 , 8/9, 16/81 and 10/27, 

HCF of given fractions = [ HCF (2,8,16,10) ] / [ LCM ( 3,9,81,27 ) ]
LCM of given fractions =  [ LCM (2,8,16,10) ] / [ HCF ( 3,9,81,27 ) ]

Let's express all the given numbers as the product of prime factors to find HCF and LCM.

2 = 21 * 50                           3 = 31
8 = 23 * 50                                 9 = 32
16 = 24 * 50                        81 = 34
10 = 21 * 51                        27 = 33

Looking at above we see :
HCF (2,8,16,10) = 21 * 50 = 2 * 1 = 2 ( To find the HCF we need to check the least power among factors)
LCM (2,8,16,10) = 24 * 51 = 16 * 5 = 80 ( To find the LCM we need to check the highest power among factors)
HCF ( 3,9,81,27 ) = 31 = 3 ( To find the HCF we need to check the least power among factors)
LCM ( 3,9,81,27 ) = 34 = 81( To find the LCM we need to check the highest power among factors)

Therefore,
HCF of given fractions = [ HCF (2,8,16,10) ] / [ LCM ( 3,9,81,27 ) ] = 2/81
LCM of given fractions =  [ LCM (2,8,16,10) ] / [ HCF ( 3,9,81,27 ) ] = 80/3
    (2) Three numbers are given: 0.63, 1.05, 2.1, We need to find their HCF and LCM.
To find the HCF or LCM of decimal numbers.
Multiply all the numbers by 10 as many times until they become integer.
All the numbers should be multiplied by 10 same number of times and then once the LCM/HCF is found then divide it by same number of 10s as multiplied.

    Here, we will multiply all the numbers by 100, as 100 will make all of them an integer. So the numbers become: 63, 105 and 210.

Let's express all the given numbers as the product of prime factors to find HCF and LCM.

63 = 71 * 32 * 50 * 20
105 = 71 * 51* 31 * 20
210 = 71 * 31 * 51 * 21
Looking at above we see :
HCF (63,105,210) = 71 * 31 * 50 * 20 = 7 * 3 *1 * 1 = 21 ( To find the HCF we need to check the least power among factors)
LCM (63,105,210) = 71 * 32 * 51 * 21 = 630 ( To find the LCM we need to check the highest power among factors)

Now, we will divide the HCF and LCM by 100 to get the final answer.
Therefore,
HCF ( 0.63, 1.05, 2.1 ) = 0.21
LCM ( 0.63, 1.05, 2.1 ) = 6.30

    (3) We know the formula:
    Product of two numbers = Product of their HCF and LCM
    Other number * 77 = ( 11 * 693 )
    Therefore,
    Other number = [ ( 11 * 693 ) ] / 77 = 99

    (4) To find Greatest possible length which can be used to measure exactly the lengths 4m 95 cm, 9m and 16m 65 cm.
    Greatest possible length will be the highest common factor of :
4m 95 cm, 9m and 16m 65 cm.
Lets first convert all of them to same unit.
4m 95 cm, 9m and 16m 65 cm = 495 cm, 900 cm, 1665 cm.
Hence the answer is HCF ( 495, 900, 1665 ) = 45 cm
( Use either factorization or division method to find HCF : http://arithmophobia-elixir.blogspot.in/2014/07/hcf-and-lcm.html )


    (5) To find the greatest numbers which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
    Let x be the number.
Therefore, when 1657 is divided by x, remainder that is left is 6 and,
when 2037 is divided by x, remainder that is left is 5,
That means (1657 – 6) and (2037 – 5) are completely divisible by x.
That means 1651 and 2032 are completely divisible by x.
Hence the greatest number x, which divides 1651 and 2032 completely is:
HCF ( 1651, 2032 ) = 127
Explanation of above HCF by long division method.
       ------------------
1651) 2032 ( 1
          1651
       -----------____________
                381 )    1651        ( 4
                            1524
                             ---------________
                                 127 )       381    ( 3
                                                381
                                            ---------
                                                   0
                                                          ____________

(6) To find the largest number of four digits exactly divisible by 12, 15, 18 and 27

Largest 4 digit number which is divisible by each of the given numbers must be divisible by LCM (12,15,18,27) = 540 
largest 4 digit number = 9999 
So, 9999/540. we get 279 as remainder. 

Therefore, t
he largest 4 digit number exactly divisible by 12,15,18,27

= 9999 – 279 = 9720 

      (7) To find the smallest number of five digits exactly divisible by 16, 24, 36 and 54
      Smallest 5 digit number which is divisible by each of the given numbers must be divisible by LCM (16, 24, 36, 54) = 432
      Smallest
      5 digit number = 10000
      So, 10000/432. we get 64 as remainder. 
      Now we need to subtract 64 to 10000 to get the smallest 5 digit number exactly divisible by 16, 24, 36, 54, but this will make the number as 4 digit number. Hence
      the four digit number (10000-64 is divisible by 432, but to make it five digit smallest number divisible by 432 we need to add 432 )
      = 10000 - 64 + 432 = 10368  
    (8) Given numbers are 99, 101, 176,182 . We need to find the number with greatest number of divisors.
    Let’s express each numbers as product of prime factors:
    99 = 3 * 3 * 11 = 32 * 111
    101 = 1011
    176 = 2 * 2 * 2 * 2 * 11 = 24 * 111
    182 = 2 * 7 * 13 = 21 * 71 * 131

Number of divisors of an integer = Product of (exponents + 1) of all the prime factors.
Therefore,
Number of divisors of 99 = (2 + 1) * (1+1) = 6
Number of divisors of 101= 1 + 1 = 2
Number of divisors of 176 = (4 + 1) * (1 + 1) = 10
Number of divisors of 182 = (1 + 1) * (1 + 1) * (1 + 1) = 8

Therefore, 176 is the number with the greatest number of divisors.


      (9) To find the number of number-pairs lying between 40 and 100 with their HCF as 15.
      Numbers with HCF 15 must contain 15 as a factor.
    multiples of 15 between 40 and 100 are; 45,60,75,90;
    HCFof 45,60 = 15;
    HCF of 45,75 = 15;
    HCF of 45,90 = 45;
    HCF of 60,75 = 15;
    HCF of 60,90 = 30;
    HCF of 75,90 = 15


    Hence, number of such pairs are : 4


    (10) To find the number of numbers pairs, whose product is 2028 and their HCF is 13
Let the numbers be 13a and 13b.
Then, 13a * 13b = 2028, or ab = 12
or, a * b = 12
or, a * b = 4 * 3 = 12 * 1
Thus, a = 12 or 4 and b = 1 or 3
Hence, the numbers are 13 *12 and 13 *1 or 13 *4 and 13 *3.
Therefore, there are 2 such pairs.









Questions on HCF & LCM



==========================================================
Before solving the questions do revise : 
Definition and Formula:
http://arithmophobia-elixir.blogspot.in/2014/07/basic-facts-and-formula.html
Methods to find HCF & LCM:
http://arithmophobia-elixir.blogspot.in/2014/07/hcf-and-lcm.html

Part-A has easier questions compared to Part-B
==========================================================

Part-A

(1) Find the HCF and LCM of 2/3 , 8/9, 16/81 and 10/27

(2) Find the HCF and LCM of 0.63, 1.05 and 2.1

(3)  The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other.

(4) Find the greatest possible length which can be used to measure exactly the lengths 4m 95 cm, 9m and
16m 65 cm.

(5) Find the greatest numbers which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

(6) Find the largest number of four digits exactly divisible by 12, 15, 18 and 27.

(7) Find the smallest number of five digits exactly divisible by 16, 24, 36 and 54.

(8) Which of the following has most number of divisors.
             (a) 99                        (b) 101                       (c) 176                        (d) 182

(9) The number of number-pairs lying between 40 and 100 with their HCF as 15 is:
             (a) 3                         (b) 4                            (c) 5                            (d) 6

(10) The product of two numbers is 2028 and their HCF is 13. The number of such pairs is :
             (a) 1                         (b) 2                            (c) 3                            (d) 4

Checkout the answers and solutions at http://arithmophobia-elixir.blogspot.in/2014/07/solutions-hcf-lcm-part-a.html

Part-B

(11) Three numbers which are co-prime to each other are such that the product of first two is 551
and that of last two is 1073. The sum of the three numbers is:
             (a) 75                       (b) 81                          (c) 85                           (d) 89

(12)  Three numbers are in the ratio of 3:4:5 and their LCM is 2400. Their HCF is:
             (a) 40                       (b) 80                          (c)120                          (d) 200

(13) The sum of two numbers is 2000 and their LCM is 21879. The two numbers are:
             (a) 1993, 7              (b) 1991, 9                  (c) 1989, 11                (d) 1987, 13

(14) The LCM of three different numbers is 120. Which of the following cannot be their HCF ?
             (a) 8                         (b) 12                           (c) 24                           (d) 35

(15) The smallest number which when diminished by 7, is divisible by each one of 12, 16, 18, 21 and
28 is :
            (a) 1008                    (b) 1015                       (c) 1022                      (d) 1032

(16) The smallest number which when increased by 5, is divisible by each one of 24, 32, 36 and
54 is:
            (a) 427                       (b) 859                         (c) 869                        (d) 4320

(17) The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same numbers of pencils is:
           (a) 91                          (b) 910                         (c) 1001                      (d) 1911

(18)  Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes how many times do they toll together?
           (a) 4                             (b) 10                           (c) 15                           (d) 16

(19)  Four different electronic devices make a beep after every 30 minutes, 1 hour, 1 hour 30 minutes and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. They will again beep together at:
           (a) 12 a.m.                    (b) 3 a.m.                     (c) 6 a.m.                      (d) 9 a.m.

(20) A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point.
After what time will they meet again at the same point ?
           (a) 26 min 18 sec         (b) 42 min 36 sec       (c) 45 min                     (d) 46 min 12 sec



Monday, 7 July 2014

HCF and LCM




To check the definition of HCF and LCM, look into the page http://arithmophobia-elixir.blogspot.in/2014/07/basic-facts-and-formula.html

Methods of Finding HCF:

  1. Factorization Method : Express each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives HCF.
Example: Find HCF of 108, 288, 360
108 = 22 * 33 , 288 = 25 * 32, 360 = 23 * 32 * 5
We can also write it as : 108 = 22 * 33 * 50 , 288 = 25 * 32 * 50, 360 = 23 * 32 * 51 ( as 50 = 1 and 51 = 5 )
Here, we have put 50 just for understanding and to make all the factors 2, 3 and 5 common among all three numbers.
Here, We have expressed 108, 288 and 360 as a product of prime factors.
We need to break all the factors until it cannot be broken further (so as to reach the prime factor).
After factorizing we see the common factors of 108, 288 and 360 are 2, 3 and 5. Now, the least power for 2 is 2, for 3 is 2 and for 5 is 0 among all three given numbers.

Hence, HCF = 22 * 32 * 50 = 4 * 9 * 1 = 36

  1. Division Method: Suppose we have to find the HCF of two given numbers. Divide the larger number by the smaller one. Now divide the divisor by the remainder . Repeat the process of dividing the preceeding number by the remainder last obtained until zero is obtained in the remainder. The last divisor is the required HCF.
Note: To find the HCF of more than two numbers. Find the HCF of any two number. And then take that HCF and another number from the given numbers and again find the HCF of those two. Keep repeating this until at the end you get one HCF.

Suppose we have to find the HCF of three given numbers.
Then HCF of [(HCF of any two) and ( the third number)] gives the HCF of three numbers.

Example: Find HCF of 513, 1134, 1215
        ____
1134 ) 1215 ( 1
            1134
__________
81 ) 1134 ( 14
          81
__________
        324
        324
_______
          0
_______

HCF of 1134 and 1215 is 81. ( If we have to find the HCF of 1134 and 1215 then 81 is the answer)

Here we have to find the HCF of three numbers 513, 1134 and 1215.
Hence required HCF = HCF of 513 and 81.

      ___
81 ) 513 ( 6
        486
______
27 ) 81 ( 3
        81
__________
       0
_______

We see the last divisor is 27.

Hence, HCF of required numbers is 27.

Methods of Finding LCM:

  1. Factorization method: Express each one of the given numbers as the product of prime factors. The product of highest powers of common prime factors gives LCM.
Example: Find LCM of 108, 288, 360
108 = 22 * 33 , 288 = 25 * 32, 360 = 23 * 32 * 5
We can also write it as : 108 = 22 * 33 * 50 , 288 = 25 * 32 * 50, 360 = 23 * 32 * 51 ( as 50 = 1 and 51 = 5 )
Here, we have put 50 just for understanding and to make all the factors 2, 3 and 5 common among all three numbers.
Here, We have expressed 108, 288 and 360 as a product of prime factors.
We need to break all the factors until it cannot be broken further (so as to reach the prime factor).
After factorizing we see the common factors of 108, 288 and 360 are 2, 3 and 5. Now the highest power for 2 is 5, for 3 is 3 and for 5 is 1 among all three given numbers.

Hence, LCM = 25 * 33 * 5 = 32 * 27 * 5 = 4320

  1. Common Division Method: Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process until no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required LCM of the given numbers.
Example: Find LCM of 513, 1134 and 1215

3 513 1134 1215
3
171
378
405
3
57
126
135
3
19
42
45


19 14 15

Here divisors are 4 times 3 at the left and undivided numbers are 19, 14 and 15 from the bottom of the above table.
Hence required LCM = 3 * 3 * 3 * 3 * 19 * 14 * 15 = 323190


Note: We have taken same set of numbers for both LCM and HCF to make you understand the difference between LCM and HCF. We will solve more questions on this later. Please let us know if you have any doubts, by posting your comments/suggestions either on facebook ( www.facebook.com/mathselixir) or here at below. You need to login with your gmail ID or open ID to post the comments.

Friday, 4 July 2014

Formula : Surds & Indices




Surds & Indices:


Surds: The irrational root of an integer.
If you can't simplify a number to remove a square root
(or cube root etc) then it is a surd. 

Example:
Number Simplifed As a Decimal Surd or
not?
√2 √2 1.4142135... Surd
√3 √3 1.7320508... Surd
√4 2 2 Not a surd
√(1/4) 1/2 0.5 Not a surd

Indices (  or Exponents ): Powers of numbers. They are defined as how many times a number multiplies by itself.

Example: 
The expression 25 is defined as follows:
We call "2" the base and "5" the index.

Laws of Indices:
When m and n are positive integers, 
  1. am × an = am + n 
  2. am / an = am – n  
  3. (am)n = amn
  4. (ab)n = anbn
  5. (a/b)n = an / bn
  6. a0 = 1
Laws of Surds:
When m and n are positive integers

  1. n a = a1/n

  2. n ab = n a * n b

  3. n(a/b) = n a / n

  4. (n a)n = a

  5. mn a = mna

  6. (na)m = nam

    Tips to simplify expressions with surds:

    In general
    • Fractions in the form  \sqrt{\frac{1}{a}} multiply top and bottom by  \sqrt{a}
    • Fractions in the form  \frac{1}{a + \sqrt{b}} multiply the top and bottom by  a - \sqrt{b}
    • Fractions in the form  \frac{1}{a - \sqrt{b}} multiply the top and bottom by  a + \sqrt{b}


      Note: For more examples: We will be using these formula in later sessions and will refer back here.






































































Thursday, 3 July 2014

Basic Facts and Formula



(a + b)² = a² + b² + 2ab
(a - b)² = a² + b² - 2ab

a² - b² = (a + b)(a - b)

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca )

(a + b)³ = a³ + 3ab(a + b) + b³
(a – b)³ = a³ + 3ab(a – b) – b³

a³ + b³ = (a + b)(a² – ab + b²)
a³ – b³ = (a – b)(a² + ab + b²)

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca )

If (a + b + c) = 0 , then a³ + b³ + c³ = 3abc

(xⁿ - aⁿ) is divisible by (x - a) for all values of n

(xⁿ - aⁿ) is divisible by (x  + a ) for all EVEN values of n
(xⁿ + aⁿ) is divisible by (x  + a ) for all ODD values of n

(1 + 2 + 3 + ... + n) = [n(n + 1 )]/2
(1² + 2² + 3² + ... + n²) = [n(n + 1)(2n + 1)]/6
(1³ + 2³ + 3³ + ... + n³) = [n(n + 1)/2]²

Dividend = ( Divisor * Quotient ) + Remainder

Factor and Multiple: If a number "a" divides another number "b" with a remainder of zero then:
a is a factor of b
b is a multiple of a

Highest Common factor (HCF) or Greatest Common Divisor (GCD)
or Greatest Common Measure (GCM):
The HCF of two or more numbers is the greatest number that divides each of them exactly.

Least Common Multiple (LCM)
The least number which is exactly divisible by each one of the given numbers is called their LCM.

Product of two numbers = Product of their HCF and LCM

To check the methods of finding LCM and HCF and examples on it check the page:
http://arithmophobia-elixir.blogspot.in/2014/07/hcf-and-lcm.html

Fraction ( or Common Fraction or Vulgar Fraction ): A number of the form a/b where the number a is not divisible by b is called fraction.
Here an and b both are integers.
Here a is called numerator and b is called denominator

HCF of Fractions = ( HCF of Numerators ) / ( LCM of Denominators )
LCM of Fractions = ( LCM of Numerators ) / ( HCF of Denominators )

Decimal Fractions: Fractions in which Denominators are powers of 10.

Decimal representation: A decimal representation of a non-negative real number r is an expression of the form of a series, traditionally written as a sum
 r=\sum_{i=0}^\infty \frac{a_i}{10^i}
where a0 is a nonnegative integer, and a1, a2, … are integers satisfying 0 ≤ ai ≤ 9, called the digits of the decimal representation. The sequence of digits specified may be finite, in which case any further digits ai are assumed to be 0.

Recurring Decimal:  The decimal representation of a number is said to be repeating if it becomes periodic (repeating its values at regular intervals) and the infinitely-repeated portion is not zero.
For example, the decimal representation of ⅓ becomes periodic just after the decimal point, repeating the single-digit sequence "3" forever, i.e. 0.333….
The repeated portion is expressed by putting a bar or a dot at its top.

Pure Recurring Decimal: A decimal representation in which all the figures after the decimal point are repeated.

Mixed Recurring Decimal: A decimal representation in which some figures do not repeat and some of them are repeated.

Representation of Recurring Decimal:
Fraction Ellipsis Vinculum Dots Parentheses
1/9 0.111… 0.1 0.\dot{1} 0.(1)
1/3 0.333… 0.3 0.\dot{3} 0.(3)
2/3 0.666… 0.6 0.\dot{6} 0.(6)
9/11 0.8181… 0.81 0.\dot{8}\dot{1} 0.(81)
7/12 0.58333… 0.583 0.58\dot{3} 0.58(3)
1/81 0.012345679… 0.012345679 0.\dot{0}1234567\dot{9} 0.(012345679)
22/7 3.142857142857… 3.142857 3.\dot{1}4285\dot{7} 3.(142857)


Converting a Pure Recurring Decimal into Vulgar Fraction:
Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures.

Converting a Mixed Recurring decimal into Vulgar fraction:
In the numerator take the difference between the number formed by all the digits after decimal ( taking repeated digits only once) and that formed by the digits which are not repeated.
In the denominator take the number formed by as many zeros as is the number of non-repeating digits.

Algebraic Expression ( or Expression ) :  An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like x or y) and operators (like add,subtract,multiply, and divide.

BODMAS rule: The correct sequence in which the operations are to be executed in order to find out the value of a given expression.
Here, (1) B = Bracket, (2) O = of , (3) D = Division, (4) M = Multiplication, (5) A = Addition, (6) S= Subtraction. In the same order of 1 to 5, we simplify an expression.
note: If an expression contains bar( or vinculum) then before applying "BODMAS" rule, simplify the expression under vinculum.

Modulus of a Real number a is defined as:
| a | = a, if a > 0
| a | = -a. if a < 0

Square root: If x² = y, we say that the square root of y is x and we write √y = x.

Cube root: The cube root of a given number x is the number whose cube is x. The cube root of x is denoted by  ³√x

Prime number: A number greater than 1 is called prime if it has exactly two factors the number 1 and the number itself.

Composite number: Numbers greater than 1 which are not prime are called composite number.

Total Prime numbers upto 100 are 25 in number.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Prime numbers greater than 100:
Let p be a given prime number greater than 100.
Method to find:
k > √p , where k is a whole number greater than square root of p
Test whether p is divisible by any prime number less than k.
Yes divisible, then p is no prime.
No not divisible, p is prime.

Example: Is 191 prime number?
Here p=191
Let's find a whole number greater than √191
Now we know, 13*13=168 and 14*14=196, that means square root of 191
is greater than 13 and less than 14.
i.e 14 > √191 (So lets's take the whole number as 14)
1.e k=14
Prime numbers less than 14 are 2, 3, 5, 7, 11 and 13
191 is not divisible by any of them.
Therefore, 191 is a prime number.

Co prime numbers: Two numbers are said to be co prime if the highest common factor of
them is the number 1.
Example: 2 and 3 are co prime numbers
               8 and 11 are co prime numbers

If a number is divisible by p as well as q, where p and q are co-primes,
then the given number is divisible by pq.
But if p and q are not co-primes then the given number need not be divisible by pq,
even if it is divisible by both p and q.

Average = ( Sum of Observations) / ( Total Number of Observations )

We may be add links to this page redirecting you to the other pages as we add more examples on above topics.

----------------------

Some of the keyboard strokes used for typing:
alt 0184 = ¼
alt 0189 = ½
alt 0175 = ¯
alt 251 = √
(alt + 0179) + (alt + 251) = ³√

alt 228 = Σ


Wednesday, 2 July 2014

Divisibility Rules on Numbers


A number is divisible by :
2  if the last digit is 0, 2, 4, 6 or 8.
3  if the sum of the digits is divisible by 3.
4  if the number formed by the last two digits is divisible by 4.
5  if the last digit is either 0 or 5.
6  if it is divisible by 2 AND it is divisible by 3.
8  if the number formed by the last three digits is divisible by 8.
9  if the sum of the digits is divisible by 9.
10  if the last digit is 0.
11 if you sum all alternate digits and then subtract this sum from the sum of the rest of the digits answer is 0 or 11.
Example: 1331 ((3+1) - (1+3) = 0). 1331 is divisible by 11.
       25176 ((5+7) - (2+1+6) = 3). 25176 is not divisible by 11.


Divisibility by 7:
This can be tested by a recursive method. A number of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7.
In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a number known to be divisible by 7 is obtained.
The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7.
For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10 = −7; thus, since −7 is divisible by 7, 371 is divisible by 7.

Another method is multiplication by 3. A number of the form 10x + y has the same remainder when divided by 7 as 3x + y.
One must multiply the leftmost digit of the original number by 3, add the next digit, take the remainder when divided by 7, and continue from the beginning: multiply by 3, add the next digit, etc.
For example, the number 371: 3×3 + 7 = 16 remainder 2, and 2×3 + 1 = 7. This method can be used to find the remainder of division by 7.