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ALGEBRA is great fun

Algebra is great fun - you get to solve puzzles!
With computer games you play by running, jumping or finding secret things. Well, with Algebra you play with letters, numbers and symbols, and you also get to find secret things!
And once you learn some of the "tricks", it becomes a fun challenge to work out how to use your skills in solving each "puzzle".equation
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how is tan (A+B) = [tan A + tan B]/[1 - tan A tan B]?

How is tan (A+B) = [tan A + tan B]/ [1 - tan A tan B]? Method 1: Let A = 30 deg and B = 45 deg. LHS = tan (30+45) = tan 75 = 3.732050808 RHS = [tan A + tan B]/[1 - tan A tan B] = [tan 30 + tan 45]/[1 - tan 30 tan 45] = [0.577350269 + 1]/[1 - 0.577350269*1] = 1.577350269/0.42264973 = 3.732050808 = LHS Proved. Method 2: tan (A+B) = [tan A + tan B]/[1 - tan A tan B] RHS = [tan A + tan B]/[1 - tan A tan B] =[(sin A/cos A) + (sin B/cos B)]/[1-(sin A/cos A)(sin B/cos B) = [sin A cos B + cos A sin B]/[cos A cos B][1 - sin A sin B/(cos A cos B)] = sin (A+B)/{[cos A cos B][cos A cos B - sin A sin B]/(cos A cos B)} = sin (A+B)/[cos A cos B - sin A sin B = sin (A+B)/cos (A+B) = tan (A+B) = LHS. Proved. Thanks. source:Quara
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What are the last 3 digit of 2^2017?

I will provide two methods for this. Method 1  (Easy way) : USE A CALCULATOR. You will get the answer in a matter of seconds like I got. Clearly, the answer is 072. Method 2  (Slight harder way) : Here, I want to find the answer without using the calculator. Let's try it. Note :- I will be making use of  Congruence Modulo  and  Euler's Theorem , so these are the prerequisites. Another way to put the question is  “Find the remainder when  2 2017 2 2017  is divided by 1000”. First of all we factorise 1000 as: 1000 = 2 3 × 5 3 1000 = 2 3 × 5 3 Next, we find the remainder by  2 3 2 3  and  5 3 5 3 seperately. It's obvious that  2 3 2 3  divides  2 2017 2 2017 . Hence, 2 2017 ≡ 0 ( m o d 8 ) 2 2017 ≡ 0 ( m o d 8 ) Now to find the remainder by 125 (or 5^3), we use Euler's Theorem. Euler's theorem is applicable in this case since  g c d ( 2 , 125 ) = 1 g c d ( 2 , 125 ) = 1 . Φ ( 125 ) = ...
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