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Question
Find the last two digits of the number 3600
Solution
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.
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