**How to Solve Maths (Shortcuts).**

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To find the number of factors of a given number, express the number as a **product** of powers of prime numbers.

In this case, 48 can be written as 16 * 3 = (2^{4} * 3)

Now, increment the power of each of the prime numbers by 1 and multiply the result.

In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)

Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 – 2 = 8 factors.

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The sum of first n natural numbers = n (n+1)/2

The sum of squares of first n natural numbers is n (n+1)(2n+1)/6

The sum of first n even numbers= n (n+1)

The sum of first n odd numbers= n^2

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To find the squares of numbers near numbers of which squares are known

To find 41^2 , Add 40+41 to 1600 =1681

To find 59^2 , Subtract 60^2-(60+59) =3481

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If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.

eg: x^4+3x^2+2x+6=0 has no positive roots .

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For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) .

Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)

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For a cubic equation ax^3+bx^2+cx+d=o

sum of the roots = – b/a

sum of the product of the roots taken two at a time = c/a

product of the roots = -d/a

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