This page describes some of the key shortcuts in maths primarily in algebra such as multiplication, square, quadratic equations and large exponents of a number. Do you kown that 2, 3, 7 ad 8 cannot be at unit's place of a perfect square, that square of a number having 'n' digits will results in a number with 2n or 2n-1 digits, that the square of any odd number is sum to two consecutive numbers and so on. That is:
5^{2} = 12 + 13
9^{2} = 40 + 41
...
13^{2} = 64 + 65
We have always been explained the standard method of finding the roots of a quadratic equation. Do you know that the first derivative of the equation = discriminant!
What is the time you will need to calculate 997×998? Using Vedic mathematics, you can simply write the product 997×998 = 995006!
Vedic mathematics contain fascinating hidden gems which are not only interesting but also help improve calculation speed. This is a complementary page where we endeavour to provide tips and tricks in mathematics. The questions dealing with standard aptitude tests are described such as finding remainder in divisions, efficiency method, total effort method, mathematical induction and so on.
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 and [b] number comprising of digits at tens and units is divisible 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
a^{2} = (a + b)*(a - b) + b^{2}. 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: d^{2}. Write the right two digits, xx and carry the third digit if any say c. Here, d^{2} → 16^{2} → 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: 84^{2} = 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. x5^{2} = x*(x+1)|25. Thus: 75^{2} = 7*8|25 = 5625, 115^{2} = 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.
a^{2} - b^{2} = (a + b) × (a - b): This equation has many applications in solving maths problems quickly. For example:
The sum of two numbers is 45 and their difference is 10. The difference of their squares = 45 × 10 = 450.
The difference between the squares of two consecutive numbers is 25. Thus: a + b = 25 and a - b = 1 and hence a = 12, b = 13.
The difference in squares of two consecutive number 50 and 51, that is 50^{2} - 51^{2} = (50 + 51) × (51 - 50) = 101.
a^{2} + b^{2} = (a - b)^{2} + 2ab. Thus, if the product of two consecutive numbers is known, the sum of their squares = twice the product + 1.
a^{2} + b^{2} = (a + b)^{2} - 2ab. Example, the sum of two numbers is 23 and their product is 126. Find the sum of squares of these numbers. a^{2} + b^{2}= 23^{2} - 2×126 = 277.
a^{3} - b^{3} = (a - b)^{3} +3ab(a - b): This equation has many applications in solving maths problems quickly. For example:
The difference between the cubes of two consecutive numbers = 3ab + 1. Thus, if the product of two onsecutive number is 380, find the difference of their cubes. The answer would be 3×380+1 = 1141.
The difference in cubes of two numbers is 14716 and difference between the two number is 4. Find the product of the numbers. a^{3} - b^{3} = 14716, a - b = 4, ab = [14716 - 4^{3}]/3/4 = 1221.
a^{4} - b^{4} = (a - b) (a + b) (a^{2} + b^{2}): For two consecutive numbers: a^{2} + b^{2} = 2ab + 1, and hence a^{4} - b^{4} = (a + b) × (2ab + 1). Example: The sum of two sonsecutive number is 15 and their product is 56. Find the difference in their 4th exponent. a^{4} - b^{4} = 15 × (2×56 + 1) = 15 × 113 = 1695.
Similarly, a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2a^{2}b^{2}: For two consecutive numbers: a^{2} + b^{2} = 2ab + 1, and hence a^{4} + b^{4} = (2ab + 1)^{2} - 2a^{2}b^{2}. Note that the sum of fourth exponent of any two consecutive number now can be evaluated from the product of these numbers.
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
1001^{2} = 1002001. Hence, 1001^{2} - 1 is divisible by 1000 or multiple of 1000 as per divisibility rules of 2, 3, 4 ...
1001^{3} = 1003003001. Hence, 1001^{3} - 1 is divisible by 1000 or multiple of 1000 as per divisibility rules of 2, 3, 4 ...
Special cases
1111...1^{2} [where '1' is repeated 'n' times] = 123..n(n-1)...1. e.g. 11111^{2} = 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: 111111111111^{2} = 12345679012320987654321, here U = 3 in (12+1 = 13).
1 repeated 13 times: 1111111111111^{2} = 1234567901234320987654321, here U = 4 in (13+1 = 14).
3333...3^{2} [where '3' is repeated 'n' times] = 11..1088..89 where '1' and '8' are repeated 'n-1' times. e.g. 333333^{2} = 111110888889 where n = 6.
666...6^{2} [where '6' is repeated 'n' times] = 44..4355..56 where '4' and '5' are repeated 'n-1' times. e.g. 6666666^{2} = 44444435555556 where n = 7.
999...9^{2} [where '9' is repeated 'n' times] = 99..9800..01 where '9' and '0' are repeated 'n-1' times. e.g. 9999^{2} = 9999800001 where n = 4.
Can you find a pattern in following square numbers formed by repetition of 2, 4, 5 and 8?
Approximate Square Root of a number
Infinite sequence of square root can be calculated as: The addition and subtraction of conjugate squares:
Similarly, last digit of 4^{n} is 6 is n is an even number and 4 if n is an odd number.
Last digit of 5^{n} is always 5.
Last digit of 6^{n} is always 6.
Last digit of 7^{n}:
7^{n} = 1 if remainder(n/4) = 0
= 7 if remainder(n/4) = 1
= 9 if remainder(n/4) = 2
= 3 if remainder(n/4) = 3
8^{n} = 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 3^{34}/5 can be easily obtained as explained below.
If Q = A × n + R, then remainder of Q^{m}/n = remainder of R^{m}/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 3^{34}/n = remainder[ { remainder of (3^{4})^{8}/5 × remainder(3^{2}/5) } / 5 ]
The trick is to break the larger exponent in smaller exponent such that remainder is 1.
Cube root of exact cubes
First digit of cube
First digit of cube root
1
1
2
8
3
7
4
4
5
5
6
6
7
3
8
2
9
9
The number of digits in a cube root = number of 3 digit groups in original cube including a single digit or double digit group if there is any.
The first (leftmost) digit of the cube root can be guessed easily from the first group (leftmost) of the cube. Thus: 17,233,489,287: there are 4 groups of 3 digits. Hence, the cube root of this number will have 4 digits. The digit at unit's place, as per the table above will be 3. The digit at thousand's place will be 2 as 2^{3} < 17.
Simple Equations
Quadratic equations in terms of receiprocals:
Same method applies when there is a minus sign between x and 1/x.
Find tens place of any exponent of a number ending in 1
E.g. tens place of 31^{57}. 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 31^{57}. 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.
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?
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 A.P. of squares
Sum of A.P. of Cubes and Products
Sum of exponents of a number and its inverse
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·x^{2} + b·x + c = 0 can be re-written as
x^{2} + b/m · x + c/m = 0
If r_{1} and r_{2} are roots of the equation, (x - r_{1}) × (x - r_{2}) = x^{2} - (r_{1} + r_{2})x + r_{1}×r_{2} = 0. Thus: r_{1} + r_{1} = -b/a and r_{1} × r_{2} = c/a.
Example:
3·x^{2} + 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. 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:
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.
Multiplications of large numbers: can you find a pattern or data is insufficient?
More symmetry in maths
1 x 8 + 1 = 9
1 x 9 + 2 = 11
12 x 8 + 2 = 98
12 x 9 + 3 = 111
123 x 8 + 3 = 987
123 x 9 + 4 = 1111
1234 x 8 + 4 = 9876
1234 x 9 + 5 = 11111
12345 x 8 + 5 = 987 65
12345 x 9 + 6 = 111111
123456 x 8 + 6 = 987654
123456 x 9 + 7 = 1111111
1234567 x 8 + 7 = 9876543
1234567 x 9 + 8 = 11111111
12345678 x 8 + 8 = 98765432
12345678 x 9 + 9 = 111111111
123456789 x 8 + 9 = 987654321
123456789 x 9 +10 = 1111111111
1 x 1 = 1
9 x 9 + 7 = 88
11 x 11 = 121
98 x 9 + 6 = 888
111 x 111 = 12321
987 x 9 + 5 = 8888
1111 x 1111 = 1234321
9876 x 9 + 4 = 88888
11111 x 11111 = 123454321
98765 x 9 + 3 = 888888
111111 x 111111 = 12345654321
987654 x 9 + 2 = 8888888
1111111 x 1111111 = 1234567654321
9876543 x 9 + 1 = 88888888
11111111 x 11111111 = 123456787654321
98765432 x 9 + 0 = 888888888
111111111 x 111111111 = 12345678987654321
Trivia
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.
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.
Aside from 1, first positive number which is square and cube of a number is 64.
365 = 10^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2}
1 googol = 1^{100} - first used in 1940 by nine-year old Mitton Sirotta
20615673^{4} = 2682440^{4} + 15365639^{4} + 18796760^{4} - integer solution for the equation of the form x^{4} + y^{4} + z^{4} = a^{4} found by Noam Elkies.
204^{2} = 23^{3} + 24^{3} + 25^{3}
16^{3} + 50^{3} + 33^{3} = 165033
Lagrange’s Four Square Theorem: every natural number can be written as sum of squares of four non-negative integers (including 0) where numbers can be repeated. The summation for few Fibonacchi numbers are:
The following integration formula have wide applications in applications such as Fourier transforms, calculation of lift coefficient of airfoils...
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)
Lateral Thinking
This is an activity where children are encouraged to find a link between arrangement of words and numbers into standard phrases, something similar to clues given in a treasure hunt. Refer to few examples below:
In the example above, ECNALG can be interpreted as Reverse Glance. Similarly, PRO / MISE can be used to refer "Broken Promise".
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.