Approximating square roots

On questions that are multiples of ten, you are able to approximate, and some of these include a square root of a relatively large number. You can easily approximate this number by taking off sets of two numbers from the end and approximating the first 3-5 numbers, then add a 0 for every 2 you took off.

Ex: √(139456)

take off the last 2, so the 56, and you're left with 1394. 35² is 1225, and 40² is 1600, and that 1394 is about in the middle, so about 37². Write it down, then add the zero from taking 2 off the end.

370 is what you would write down.

Ex: √(8675309)

if you take off the last 4 digits, you are left with 867. 25² = 625, and 30² is 900. 867 is closer to 30².
Then you would write down 29, and then add 2 zeroes (for the 4 numbers we took off).
So write down 2900.

Subtracting Reverses

There are two forms of this trick.
1. 3-digit reverses(753 - 357)
Subtract the 1st digit of the 1st number and the 1st digit of the 2nd number
7 - 3 = 4
Then plug your answer into the equation: 100n - n to get the final answer.
100(4) - 4 = 396
753 - 357 = 396
2. 4-digit reverse pairs(7568 - 6875)
Subtract the 1st pair from the second pair.
75 - 68 = 7
Then plug your answer into the equation: 100n-n to get the final answer.
100(7) - 7 = 693
7568 - 6875 = 693

Adding a Sequence in the Form 1 + 3 + .... + 2n - 1

When you have a sequence in the format similar to:

1 + 3 + 5 + 7 + ... + 2n - 1

1) Take the last number presented and add it to 1.

2) Divide the number by 2.

3) Square the number.

You now have the total of the sequence!

Ex: 1 + 3 + 5 + ... + 27

1) 27 + 1 = 28

2) 28/2 = 14

3) 14² = 196

Therefore, 1 + 3 + 5 + ... + 27 = 196

Ex: 1 + 3 + 5 + ... + 55

1) 55 + 1 = 56

2) 56/2 = 28

3) 28² = 784

Now try:
Ex: 1 + 3 + 5 + ... + 15

Ex: 1 + 3 + 5 + ... + 39

Ex: 1 + 3 + 5 + ... + 31