This video demonstrates how to solve complex exponential expressions by recognizing algebraic patterns. The problem 2^18 + 6^10 + 3^20 is solved by rewriting it as (2^9)^2 + 2(2^9)(3^10) + (3^10)^2, which matches the perfect square identity a² + 2ab + b² = (a+b)². This simplifies to (2^9 + 3^10)², and taking the square root gives 2^9 + 3^10 = 512 + 59049 = 59561. The key insight is recognizing when exponential expressions can be transformed into algebraic identities to simplify complex calculations.
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Solving a 'Harvard' University entrance exam questionAdded:
Welcome to SmartMath. Today, I'm solving this math Olympic problem without using a calculator. So, let's find out the solution.
So, our equation is 2 to the power of 18 plus 6 to the power of 10 plus 3 to the power of 20 square root. So, this means that 2 if I'm 18 is 9 into 2 plus 6 is uh if I'm take 6 and 2 10 plus uh 20 is 3 into 10 into 2 2s are 20.
Square root.
So, this means that if I'm taking 2 into 9 into 2 plus if I'm take 6 is 2 into 3 is 6 power of 10 plus 3 10 into 2 square root.
So, as I saw saw the equation is a to the power of m into n that is equal to a to the power of m whole power n.
So, this means that a {comma} b of x a into b if a into b of whole power so power is separate and with x.
So, now as we saw the equation is 2 2 2 to the power of 9 whole power 2 plus 2 to the power of 10 into separate power 3 to the power of 10.
So, plus 3 to the power of 10 whole square.
Square root.
So, this term is means that uh square root nine uh two to the power of nine nine squared plus if I'm take two uh power 10 is one plus nine into three is power three power 10 plus three to the power of 10 whole squared So, this means that our equation will be two to the power of nine whole squared plus two to the power of one into two to the power of nine into three to the power of 10 plus three to the power of 10 whole squared square root So, this means that this look like the formula of a squared a squared plus two into a into b plus b squared that is equals to a squared that is equals to a plus b whole squared So, our equation will be a rather a is two a and b is three to the power of 10 So, our above equation will be two to the power of nine plus three to the power of 10 whole squared So, uh square root and square cancelled out So, our remainder term is two to the power of nine plus three to the power of 10 So, if I'm taking uh if I'm wrote the nine is three into three is nine plus three 10 is two into 5 is 10.
So, if I am whole power 3 take whole power 3 plus and taken 3 2 whole power 5.
So, 2 into 2 into 2 is 8 cube plus and 3 and 3 2 3 into 3 is 9 whole power 5.
So, this is 8 8 cube is 8 cube is 512 plus 9 9 power of 5.
Or I'm going there.
9 to the power of 5 plus 512.
Now, if I am add and subtract 9, the equation will be same.
So, our equation will be same. This means that 9 to the power of 5 is 4 plus 1 is 5 plus if I am add these two done, they give me 521 minus 9.
So, now power is added means we separate out their basic 9 to the power 4 into 9 to the power 1 minus 9 plus 500 and 521.
So, if I'm taking common 9 here so 9 here, remainder is 9 to the power 4 minus 1 and plus 521.
So, if I'm going there 9 is to the 9 to the power 4 is 9 in 9 squared into whole square minus 1 plus 521.
This 1 is equals to 1 1 is equals to 1 square. So, this time is squared.
So, using the formula of a squared minus b squared, that is a plus b into a minus b.
That is equal to 9 into 9 squared minus 1 into 9 squared plus 1 plus 521.
21.
So, 9 squared is 81 into 81 minus 1 81 plus 1 plus 521.
So, 9 is and 81 minus 1 is 80 into 81 plus 1 is 82 plus 521.
So, 9 is if I'm So, if I'm wrote that 9 is into 80 and 80 is 80 plus 2 82 plus 521.
So, now if I multiply 9 9 into uh if I'm now if I multiply 80 into 8s are that is 8 8s are 6400 plus 82s are 160 plus 521.
So, this time is if I'm add this this time 9 is 6560 plus 521.
If 9 is 10 minus 1 and 6560 plus 521.
So, 65 10s are 65 600 minus 65 60 plus 521 So, if I'm subtract these two time, they give me 590 40 plus 521 that is 500 and 9 59 561 is our desired answer.
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