Question

Find the last two digits of the number 3⁶⁰⁰ 

Answer 1

Consider 3⁶⁰⁰

3⁶⁰⁰ = (3²)³⁰⁰

= 9³⁰⁰

= (10 – 1)³⁰⁰ 

(10 – 1)³⁰⁰ = `""^300"C"_0(10)^300 * (- 1)^0 + ""^300"C"_1 (10)^(300 - 1) * (- 1)^1 + .... + ""^300"C"_299 (10)^1 + ""^300"C"_300 (10)^0 (- 1)^300`

= 10³⁰⁰ – 300 (10)²⁹⁹ + ……………. + 300 C₁ × 10 × – 1 + 1 × 1 × 1

= 10³⁰⁰ – 300 (10)²⁹⁹ + …………….. – 300 × 10 + 1

= 10³⁰⁰ – 300 × 10²⁹⁹ + …………… – 3000 + 1

All the terms except the last are multiples of 100

And hence divisible by 100.

∴ The last two digits will be 01.