Advertisements
Advertisements
Question
Find the last two digits of the number 3600
Advertisements
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.
APPEARS IN
RELATED QUESTIONS
Evaluate the following using binomial theorem:
(101)4
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
Find the 5th term in the expansion of (x – 2y)13.
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
Find the term independent of x in the expansion of
`(x - 2/x^2)^15`
The last term in the expansion of (3 + √2 )8 is:
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
Compute 1024
Compute 994
Compute 97
Using binomial theorem, indicate which of the following two number is larger: `(1.01)^(1000000)`, 10
Find the constant term of `(2x^3 - 1/(3x^2))^5`
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)n are equal
If a and b are distinct integers, prove that a − b is a factor of an − bn, whenever n is a positive integer. [Hint: write an = (a − b + b)n and expaand]
Prove that `"C"_0^2 + "C"_1^2 + "C"_2^2 + ... + "C"_"n"^2 = (2"n"!)/("n"!)^2`
Choose the correct alternative:
The value of 2 + 4 + 6 + … + 2n is
