Basic Mathematics for Everyone
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.
Multiplication by 9
Multiplication by 11
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].
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:
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.
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 casesFor 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).
Exception: 1 repeated 10 times: 11111111111^{2} = 1234567900987654321.
7^{n} = 1 if remainder(n/4) = 0
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.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.
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.
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.
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]. 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.
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.
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 |
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.
Integral of Trigonometric Squares
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: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.
Copyright © 2017 - All Rights Reserved - CFDyna.com
Template by OS Templates