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Basic Mathematics for Everyone

Maths Tips and Trick incl. Vedic Mathematics

This is a complementary page where we endeavour to provide some mathematics and problem solving tips and tricks. The questions deal with standard aptitude test describing methods such as efficiency method, total effort method, mathematical induction and so on.

Maths book on Coordinate Geometry by S. L. Loney
Trigonometry by S. L. Loney Trigonometry by Hall & Frink
Maths - NCERT Class IX Maths - NCERT Class X
Maths - NCERT Class XI Maths - NCERT Class XII-A
Maths - NCERT Class XII-B -

Division Rules

  • Divisibility by 2: A number is divisible by 2 if digit at unit place is even.
  • Divisibility by 3: A number is divisible by 3 if sum of all digits is also divisible by 3. E.g. 1323 is divisible by 3 because (1 + 3 + 2 + 3) = 9 which is divisible by 3.
  • Divisibility by 4: A number is divisible by 4 if number comprising of digits at tens and units is divisible by 4. E.g. 123456 is divisible by 4 because 56 is divisible by 4.
  • Divisibility by 5: A number is divisible by 5 if digit at unit place is either 0 or 5.
  • Divisibility by 6: A number is divisible by 6 (2 x 3) if [a] digit at unit place is an even number - divisibility by 2 and [b] sum of all digits is divisible by 3 - divisibility by 3.
  • Divisibility by 8: A number is divisible by 8 if number comprising of digits at hundreds, tens and units is divisible by 4. E.g. 123456 is divisible by 8 because 456 is divisible by 8.
  • Divisibility by 9: A number is divisible by 9 if sum of all digits is also divisible by 9. E.g. 1323 is divisible by 9 because (1 + 3 + 2 + 3) = 9 which is divisible by 9. The numerator is repeated, decimal place = number of digits less by 1. Example:
    Step-1: 513/9 = 513 513 513 513 repeat the numerator.
    Step-2: Location of decimal place = 2 (no of digits in 513) - 1. Thus, 513/9 = 51.3513
  • Divisibility by 11: A number is divisible by the difference in sum of digits at alternate place is divisible by 11. E.g. 7238561 is divisible by 11 because [1+5+3+7] - [6+8+2] = 0 which is divisible by 11.
  • Divisibility by 12: A number is divisible by 12 (3 x 4) if [a] sum of all digits is also divisible by 3 - divisibility by 3 and [b] number comprising of digits at tens and units is divisible by - divisibility by 4.


Multiplication by 9

Multiplication by 11


Number of digits in products

Product of integers with N and M digits
N = 2, M = 3: 10 * 100   =   1000,   D = 4 = [N + M - 1]
N = 2, M = 3: 99 * 100   =   9900,   D = 4 = [N + M - 1]
N = 2, M = 3: 10 * 999   =   9990,   D = 4 = [N + M - 1]
N = 2, M = 3: 99 * 999   = 98901,   D = 5 = [N + M]

N = 2, M = 4: 10 * 1000 =   10000, D = 5 = [N + M - 1]
N = 2, M = 4: 99 * 1000 =   99000, D = 5 = [N + M - 1]
N = 2, M = 4: 10 * 9999 =   99990, D = 5 = [N + M - 1]
N = 2, M = 4: 99 * 9999 = 989901, D = 6 = [N + M]
Thus: the answer is [N + M - 1] and [N + M].


Square of a number

a2 = (a + b)*(a - b) + b2. For example:
43 * 43 = (43+3)*(43-3) + (3*3) = (46*40)+9 = 1849.
A great improvisation of this formula as per Vedic mathematics is as follows: "Whatever the extent of its difficiency, lessen it still further to that very extent; and also set-up the square of that deficiency". The calculations steps are then defined as:

  • Take the power of (10, 100, 1000 ...) nearest to the number being squared. E.g. n = 84 has base number b = 100.
  • Calculate deficiency, d = b - n = 100 - 84 = 16
  • Calculate square of the deficiency: d2. Write the right two digits, xx and carry the third digit if any say c. Here, d2 → 162 → xx = 56, c = 2
  • Subtract twice of the defficienty from base and add the carry c. Thus: aa = b - 2 * d + c. Here: a = 100 - 2 * 16 + 2 = 70.
  • Thus: 842 = aa|xx = 70|56 = 7056
  • Note that this method is best suitable if the difficiency from suitable base (10, 100, 1000 ...) is in single digit. Some adapatation will be presented later.
  • Followings are the specific applications of this rule:
    • Square of any number ending is 5: eg. x52 = x*(x+1)|25. Thus: 752 = 7*8|25 = 5625, 1152 = 11*12|25 = 13225
    • Square of numbers, when the sum of digits at unit's place is 10 and other digits are same, can be calculated using similar method. That is xy * xq where y + q = 10. xy * xq = x * (x+1)|p*q. For example: 67 * 63 = 6*(6+1)|7*3 = 4221.

      Product when sum of units place is 10

Note: 24 raised to an even power always ends with 76, 24 raised to an odd power always ends with 24, 76 raised to any power ends with 76.


Large Exponent of a Large Number

10012 = 1002001.      Hence, 10012 - 1 is divisible by 1000 or multiple of 1000 as per divisibility rules of 2, 3, 4 ...

10013 = 1003003001. Hence, 10013 - 1 is divisible by 1000 or multiple of 1000 as per divisibility rules of 2, 3, 4 ...

Special cases
  • 1111...12 [where '1' is repeated 'n' times] = 123..n(n-1)...1. e.g. 111112 = 123454321 where n = 5. This is valid till n ≤ 9.

    For n ≥ 10: 111...1 2 = 12...79012..U..2098...1 where U = unit place of (n+1). For example, 1 repeated 12 times: 1111111111112 = 12345679012320987654321, here U = 3 in (12+1 = 13).

    1 repeated 13 times: 11111111111112 = 1234567901234320987654321, here U = 4 in (13+1 = 14).

    Exception: 1 repeated 10 times: 111111111112 = 1234567900987654321.

  • 3333...32 [where '3' is repeated 'n' times] = 11..1088..89 where '1' and '8' are repeated 'n-1' times. e.g. 3333332 = 111110888889 where n = 6.
  • 666...62 [where '6' is repeated 'n' times] = 44..4355..56 where '4' and '5' are repeated 'n-1' times. e.g. 66666662 = 44444435555556 where n = 7.
  • 999...92 [where '9' is repeated 'n' times] = 99..9800..01 where '9' and '0' are repeated 'n-1' times. e.g. 99992 = 9999800001 where n = 4.
  • Can you find a pattern in following square numbers formed by repetition of 2, 4, 5 and 8?

    square of larger numbers


Approximate Square Root of a number

Square Root

Infinite sequence of square root can be calculated as:

Infinite sequence Square Root

The addition and subtraction of conjugate squares:

Conjugate square root sum


Last digit in power (exponent) of 3, 4, 7 ...

  • 34 = 81, 35 = 243, 36 = 729, 37 = 2187, 38 = 6561
    Thus:
    3n = 1 if remainder(n/4) = 0
      = 3 if remainder(n/4) = 1
      = 9 if remainder(n/4) = 2
      = 7 if remainder(n/4) = 3
  • Similarly, last digit of 4n is 6 is n is an even number and 4 if n is an odd number.
  • Last digit of 5n is always 5.
  • Last digit of 6n is always 6.
  • Last digit of 7n:

    7 power n

    7n = 1 if remainder(n/4) = 0
      = 7 if remainder(n/4) = 1
      = 9 if remainder(n/4) = 2
      = 3 if remainder(n/4) = 3
  • 8n = 6 if remainder(n/4) = 0
      = 8 if remainder(n/4) = 1
      = 4 if remainder(n/4) = 2
      = 2 if remainder(n/4) = 3
  • Using the remainders described above, remainder of expressions like 334/5 can be easily obtained as explained below.

    If Q = A × n + R, then remainder of Qm/n = remainder of Rm/n.

    If Q = (A × n + R) and P = (B × n + S), then remainder of (Q * P)m/n = remainder of (R * S)/n.

    Thus: remainder of 334/n = remainder[ { remainder of (34)8/5 × remainder(32/5) } / 5 ]

    Remainder of large exponent

    The trick is to break the larger exponent in smaller exponent such that remainder is 1.

Find tens place of any exponent of a number ending in 1

E.g. tens place of 3157. The unit place would be 1. The tens place would be unit place of product (3 * 7) and hence 1 will be at tens place of 3157. The number is calculated at product of digit at tens place in the base that is '3' here and the first digit (at units place) in the exponent which is '7' here.

So far easy for exponents. How to find digit at ten's place in multiplication of 3 or 4 digits? Refer to the method below for product 24 × 37 × 68 × 94.

Digit at tens place


Work and time - Use efficiency as short-cut

If A does the task in 20 days and B does it in 16 days. How long will they take it to complete the task together?

Efficiency Method A's efficiency = 100/20 = 5%, B's efficiency = 100/16 = 6.25%, (A+B)'s efficiency = 5+6.25 = 11.25 %. Thus: Time required = 100/11.25 = 8.89 days

Another variant of such problem is given as:
If a group-A of 4 persons each of equal efficiency can do a task in 12 days. How many day would be required to do the same task by group-B of 6 persons with efficiency 2 times that of group-A?

Equality of total effort Let x be the efficiency of persons in group A. Total effort required = 4 * x * 12 = 48x [mandays]. This is constant for that particular task.

Now, let N be the number of days required for persons in group-B to complete the task. Thus: N * 2x * 6 = 48x. Hence, N = 4 days.


Sum of Arithmetic progression of numbers and sqaures

Arithmetic Progression

Sum of Arithmetic Progression

Sum of A.P. of squares

Sum of Arithmetic Progression of Squares

Sum of A.P. of Cubes and Products

Sum of Arithmetic Progression of Cubes


Sum of exponents of a number and its inverse

Sum of exponents of a number and its inverse 01

Sum of exponents of a number and its inverse 02

Sum of exponents of a number and its inverse 03

Quadratic Equations
The method to find roots of a quadratic equation is well known. However, there are times when the standard textbook method can be shortened. For example: a·x2 + b·x + c = 0 can be re-written as

x2 + b/m · x + c/m = 0

If r1 and r2 are roots of the equation, (x - r1) × (x - r2) = x2 - (r1 + r2)x + r1×r2 = 0. Thus: r1 + r1 = -b/a and r1 × r2 = c/a.

Example:

3·x2 + 2·x - 48 = 0, the shortcut is to find two numbers which adds to -2 and whose multiplication = -48. 6 and -8 are the numbers and hence the roots of the quadratic equation are [6/3, -8/3].

Fractions

There are some general rules of the fractions which should be remembered to get answers quicker.

fraction addition to numerator and demoninator

A quick check during the aptitude test is to use: [a = 1, b = 5, c = 1] or [a = 1, b =2, c = 1].

Puzzles

There are two containers having equal volume of A and B. Now, amount x of liquid A from container is taken out and mixed with liquid B. Thereafter, same volume x is taken out from second container and mixed with liquid A. Question is: which of the containers have higher impurity? In other words, is the volume fraction of B in A in the first container is higher, lower or same as volume fraction of A in B in the second container?
Answer: The volume fraction of B in A in container 1 = volume fraction of A in B in container 2. Before the answer is demonstrated through complex calculation, the easiest way to get to the answer is by noticing the variable 'x'. It can be between 0 to 100. Hence, assume the case when x = 100 and the answer is straightforward.

Now mathematics is as follows:
Mixing of Liquids

What is the shortest path between two diagonally opposite vertices of a cube when the travel path has to be all along the walls, ground and ceiling of the cube? Two such set of vertices have been shown by circles and diamonds in the following figure.


Cube Shortest Path

Multiplications of large numbers: can you find a pattern or data is insufficient?

mutliplicationOnesFivesEights

Trivia

Other than the trivial examples of 0 and 1, the only natural numbers that equal the sum of the cubes of their digits are 153, 370, 371, and 407.

Kaprekar’s Constant - Take any four-digit number except an integral multiple of 1111 (i.e., don’t take one of the nine numbers with four identical digits). Rearrange the digits of your number to form the largest and smallest strings possible. That is, write down the largest permutation of the number, the smallest permutation (allowing initial zeros as digits), and subtract. Apply this same process to the difference just obtained. Within the total of seven steps, you always reach 6174. At that point, further iteration with 6174 is pointless: 7641–1467 = 6174. Example: Start with 8028. The largest permutation is 8820, the smallest is 0288, and the difference is 8532. Repeat with 8532 to calculate 8532–2358 = 6174. Your own example may take more steps, but you will always reach 6174.


Multiply your Money

If you invest money which compounds annualy:
  • (Approx.) number of years required to double the money = 72/ROI where ROI = annual rate of interest. E.g. if ROI = 4%, YTD (years to double) = 72/4 = 18 years. YTD varies as 70/ROI (ROI = 2%) to 75/ROI (ROI = 17%). Exact value = ln(2) / ln(1 + ROI/100).
  • (Approx.) number of years required to triple the money = 115/ROI. E.g. if ROI = 4%, YTT (years to triple) = 115/4 = 24 years. YTT varies as 111/ROI (ROI = 2%) to 120/ROI (ROI = 19%). Exact value = ln(3) / ln(1 + ROI/100)
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The content on CFDyna.com is being constantly refined and improvised with on-the-job experience, testing, and training. Examples might be simplified to improve insight into the physics and basic understanding. Linked pages, articles, references, and examples are constantly reviewed to reduce errors, but we cannot warrant full correctness of all content.