sandesh10 wrote:

Find the greatest integer

a) \(10^{10} + 2^{100}\)

b) \(100^{10} + 2^{10}\)

c) \((100 + 2 ) ^{10}\)

Can anyone suggest me the efficient way to solve this question?

Hi...

The easiest way would be..

Compare a and B..

a) \(10^{10} + 2^{100}\)

b) \(100^{10} + 2^{10}= (10^{10})^{10}+2^{10}=10^{100}+2^{10}\)

Now when you compare two 10^{100} is way GREATER than 2^{100} as compared to 10^{10} and 2^{10} ..

Otherwise a exact way would be...

a) \(10^{10} + 2^{100}=10^{10}+(2^{10})^10=10^{10}+1024^{10}=10^{10}+10^{30}~10^30\)

b) \(100^{10} + 2^{10}=10^{100}+10^3\)

SO B is clearly the bigger one

Now let's compare B and C

c) \((100 + 2 ) ^{10}\)

When you expand it 100^{10}+100^9*2^1+.....+2^10

So here C is bigger..

Also

b) \(100^{10} + 2^{10}~100^{10}\)

c) \((100 + 2 ) ^{10}=102^{10}\)

C wins

100^10+2^10=> (10^2)^10+2^10=>10^20+2^10.