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 know that 2, 3, 7 and 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. यह पृष्ठ गणित में मुख्य रूप से बीजगणित में कुछ प्रमुख संक्षिप्त रूप जैसे गुणन, वर्ग, द्विघात समीकरण और किसी संख्या के बड़े घातांक का वर्णन करता है। क्या आप जानते हैं कि 2, 3, 7 और 8 पूर्ण वर्ग की इकाई के स्थान पर नहीं हो सकते हैं, 'n' अंकों वाली संख्या के उस वर्ग का परिणाम 2n या 2n-1 अंकों वाली संख्या में होगा, जो कि किसी भी विषम संख्या का वर्ग होगा। संख्या दो क्रमागत संख्याओं का योग है और इसी प्रकार आगे भी।
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!
वैदिक गणित में आकर्षक छिपे हुए रत्न शामिल हैं जो न केवल दिलचस्प हैं बल्कि गणना की गति को बेहतर बनाने में भी मदद करते हैं। यह एक पूरक पृष्ठ है जहां हम गणित में सलाह और चाल प्रदान करने का प्रयास करते हैं। मानक अभिक्षमता परीक्षणों से संबंधित प्रश्नों का वर्णन किया गया है जैसे कि विभाजनों में शेषफल खोजना, दक्षता विधि, कुल प्रयास विधि, गणितीय प्रवर्तनआदि। 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.
* * * * * * * * * *New orientation:
* * * * * * * * * *Hint: Use the power of symmetry and you need to avoid any change to row 3. समरूपता की शक्ति का उपयोग करें और आपको पंक्ति 3 में किसी भी परिवर्तन से बचें।
(n+1)^{2} - (n+1) = n × (n+1). Thus: 7^{2} - 7 = 42 = 6 × 7. 8^{2} - 8 = 56 = 7 × 8.
64^{2}- 46^{2} = (36-16)x100 - (36-16) = 1980. इस सूत्र के अनुसार विशेष परिस्थितियों में इतनी त्वरित गणना संभव है: Such quick calculations of special cases are possible as per the formula: (a.n+b)^{2} - (b.n+a)^{2} = (a^{2} - b^{2}).n^{2} - (a^{2} - b^{2})
Do you find numbers 7, 11 and 13 boring? Not really, for any number abc x 7 x 11 x 13 = abcabc. For example, 369 x 7 x 11 x 13 = 369369. 963 x 7 x 11 x 13 = 963963.
If n is any digit between 1 and 9, nnn/3/n = 37. E.g. 666/18 = 777/21 = 37...
Multiplication by 9 - 9 से गुणा
Multiplication by 11 - 11 से गुणा
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].
Puzzle: Turn the fish by moving exactly 3 sticks
/ / \ / \ / \ \
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 deficiency, lessen it still further to that very extent; and also set-up the square of that deficiency". The calculations steps are then defined as:
Trivia: the number of dots on dices are not put randomly. They are placed such that the numbers on opposite faces add up to '7'. पासों पर बिंदुओं की संख्या यादृच्छिक रूप से नहीं डाली जाती है। उन्हें इस तरह रखा जाता है कि विपरीत फलकों पर संख्याओं का योग '7' हो जाता है।
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.
What is digit at unit's place of the number 79235168^{1/5}? The answer is 8. This is due to the fact that any number raised to its fifth power will have its last digit (that at unit's place) unchanged!
Puzzle with Numbers
Can you write a mathematical expression using 3 equal digits that results in value of 30? क्या आप 3 समान अंकों का उपयोग करके एक गणितीय व्यंजक लिख सकते हैं जिसके परिणामस्वरूप 30 का मान हो?Answer: 33 -3 = 30, 3^{3} + 3 = 30, 6 × 6 - 6 = 30, 5 × 5 + 5 = 30. Same way similar questions can be framed such as 22 + 2 = 24, 44 - 4 = 40.
Can you write a mathematical expression using 3 equal digits that results in value of that digit itself? क्या आप 3 समान अंकों का उपयोग करके एक गणितीय व्यंजक लिख सकते हैं जिसके परिणामस्वरूप उस अंक का मान स्वयं वो अंक हो?Answer: Few trivial solutions are n + n - n = n or n × n / n = n or nth square root of n^{n}. Only solution which is not trivial is 2 x 2 - 2 = 2.
Trick with Numbers
How quickly can you solve: 21^{3} - 21^{2}? As it is difficult for most of the person to calculate 21 × 21 × 21 with first calculating 21 x 21, it is time consuming. If you note that (n+1)^{3} - (n+1)^{2} = n^{3} + 2n^{2}+n, 21^{3} - 21^{2} = 20^{3} + 2 × 20^{2} + 20 = 8820.
Similarly, (n+1)^{3} - (2n+1)^{2} = n^{3} - n^{2} - n can be used to solve quickly without any complex multiplication, that is using only mental calculation of squares and cubes: 21^{3} - 41^{2} = 20^{3} - 20^{2} - 20 = 7580.
Yet another short-cut is (n+1)^{3} - (3n+1)^{2} = n^{3} - 6n^{2} - 3n can be used to solve 21^{3} - 61^{2} = 20^{3} - 6 × 20^{2} - 3 × 20 = 5540.
Another special case where the difference in cube of two consecutive numbers are to be calculated: (n+b)^{3} - (n+c)^{3} = [b-c] x [3n^{2} + 3n.(b+c) + (b-c)^{2} + 3bc]. Thus, if b-c = 1, (n+b)^{3} - (n+c)^{3} = 3n^{2} + 3n.(b+c) + 3bc + 1. Thus 73^{3} - 72^{3} = 3 x 4900 + 210 x 5 + 18 + 1 = 15769. If if b-c = 2, (n+b)^{3} - (n+c)^{3} = 2 x [3n^{2} + 3n.(b+c) + 3bc] + 2^{3}. If if b-c = 3, (n+b)^{3} - (n+c)^{3} = 3 x [3n^{2} + 3n.(b+c) + 3bc] + 3^{3} and so on.
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.
Trivia: Magic Square: A magic square is an arrangement of numbers in square shape consisting of the distinct positive integers N, N+1, N+2... (N = 1, 2, 3...) arranged such that the sum of the numbers in any horizontal, vertical or main diagonal line is always the same number. As the definition suggests, the number of rows and columns have to be equal. E.g.
4 9 2 3 5 7 8 1 64x4 square:
5 15 16 2 10 8 7 13 6 12 11 9 17 3 4 14
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.इसी प्रकार, किसी संख्या के वर्ग के वर्ग (किसी संख्या की चौथी शक्ति) में इकाई के स्थान पर या तो 1, 5 या 6 होता है। यानी यदि इकाई का स्थान एक विषम संख्या (5 को छोड़कर) है, तो उस संख्या का चौथा घातांक हमेशा 1 पर समाप्त होगा। इसी तरह, यदि इकाई का स्थान एक सम संख्या है, तो उस संख्या का चौथा घातांक हमेशा 6 में समाप्त होगा। Similarly, the square of square of a number (4th power of a number) has either 1, 5 or 6 at units place. That is if unit's place is an odd number (except 5), the 4th exponent of that number will always end in 1. Similarly, if unit's place is an even number, the 4th exponent of that number will always end in 6.
A combination of this the above rules can be used to find digit at unit's place for any other exponent (such as 6th, 8th, 10th ...) of a number. Use the following table to check the rules described thus far.Number | X^{2} | X^{3} | X^{4} | X^{5} | X^{6} |
11 | 121 | 1331 | 14641 | 161051 | 1771561 |
12 | 144 | 1728 | 20736 | 248832 | 2985984 |
13 | 169 | 2197 | 28561 | 371293 | 4826809 |
14 | 196 | 2744 | 38416 | 537824 | 7529536 |
15 | 225 | 3375 | 50625 | 759375 | 11390625 |
16 | 256 | 4096 | 65536 | 1048576 | 16777216 |
17 | 289 | 4913 | 83521 | 1419857 | 24137569 |
18 | 324 | 5832 | 104976 | 1889568 | 34012224 |
19 | 361 | 6859 | 130321 | 2476099 | 47045881 |
Quiz Time: A tree grows such that its height doubles every fourth day. If its height was 50 [m] on sixty-fourth day, when was its height 25 [m]?
Cube root of exact cubes - सटीक घनों का घनमूल
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. Alternatively, except 4, 6 and 9, the digit at unit place in = 10 - the digit.
Trivia: The following three consecutive numbers are the lowest that are divisible by cubes other than 1: 1375, 1376 and 1377 - (divisible by the cubes of 5, 2 and 3 respectively).
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.
Trivia: Tower of Hanoi - Excertps from [mathworld.wolfram.com/TowerofHanoi.html] This puzzle was invented by E. Lucas in 1883. It is also known as the Tower of Brahma puzzle and appeared as an intelligence test for apes in the film Rise of the Planet of the Apes (2011) under the name "Lucas Tower".
Puzzle: 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.
(6a^{2} - 4ab + 4b^{2})^{3} = (3a^{2} + 5ab - 5b^{2})^{3} + (4a^{2} - 4ab + 6b^{2})^{3} + (5a^{2} - 5ab - a^{2})^{3}
(a+m^{2}n)^{3} + (mb + n)^{3} = (ma + n)^{3} + (b + m^{2}n)^{3}
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 annually:What is the next two letters in this sequence: B C D G J ... ...? The answers are: O and P. Why and how?
How many straight lines are required to connect 16 dots arranged in a 4 × 4 square, the lines must not cross each other?
One of the quickest answer is: 7 as shown below. Can you find a better way?
Move 3 sticks to show only 2 squares instead of 3 shown below.
O--------O--------O | | | | | | O--------O--------O | | | | O--------O
There are two ropes which takes total 1 hour to burn. These ropes may burn at different rates - but total time remains 1 hour. You have enough matchsticks to burn them as you need. How can you burn them to measure 45 minutes? Hint: Note that any special solution is part of a more generic solution!
Can we have two right angles in a triangle? The classical textbook thinking will lead to answer in negative. That is because we always think triangles or rectangles to be formed by connecting ends of straight lines! What if the lines are not straight? Refer the image below and revisit your answer.
Note that the formula is applicable to both the situation shown in the image. In case chord location is above R, the expression (R-H) will negative and the two areas A_{1} and A_{2} will add up. Similarly, the angle θ will be calculated to be > 90°.
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