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Tips and Shortcuts 1 year 1 month ago #1

  • tasneem
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Number of Factors:
To find the number of factors of a given number, express the number as a product of powers of prime numbers.
In this case, 120 can be written as 8 x 3 x 5 = ( 2^3) (3) (5)

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

In this case it will be (3 + 1)*(1 + 1)*(1+1) = 4 * 2 * 2 = 16 (the power of 2 is 3, the power of 3 is 1 and the power of 5 is 1)

Therefore, there will 16 factors including 1 and 120. Excluding, these two numbers, you will have 16 – 2 = 14 factors.
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Sum on n Natural numbers 1 year 1 month ago #2

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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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Re: Sum on n Natural numbers 1 year 1 month ago #3

  • mitesh
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Some Pythagorean triplets: (easy way to remember them)
3,4,5
(3^2=4+5)
5,12,13
(5^2=12+13)
7,24,25
(7^2=24+25)
8,15,17
(8^2 / 2 = 15+17 )
9,40,41
(9^2=40+41)
11,60,61
(11^2=60+61)
12,35,37
(12^2 / 2 = 35+37)
16,63,65
(16^2 /2 = 63+65)
20,21,29
(EXCEPTION)
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