Number theory problem- 11 Nov

Question

!!A = (99^50 + 100^50) and B = 101^50!!, the larger is?

!!a)" "A!!

!!b)" "B!!

!!c)!!Both are same

!!d)!!Cant Say

Solution

We can solve this problem using binomial concepts.

!!A = (99^50 + 100^50) = (100-1)^50 + 100^50!!

and !!B = 101^50 = (100+1)^50!!

!!B-A = (100+1)^50 - ((100-1)^50 + 100^50)!!

!!B-A = 100^50 + 50C_1(100)^49 + 50C_2(100)^48 .......... - (100^50 - 50C_1(100)^49 + 50C_2(100)^48 - 50C_3(100)^47 .......... + 100^50)!!

!!B-A = 50C_1(100)^49 + 50C_2(100)^48 .......... - (100^50 - 50C_1(100)^49 + 50C_2(100)^48 - 50C_3(100)^47 ..........)!!

!!B-A = 50C_1(100)^49 + 50C_2(100)^48 .......... - 100^50 + 50C_1(100)^49 - 50C_2(100)^48 + 50C_3(100)^47 ..........!!

!!B-A = 2xx50C_1(100)^49 + 50C_2(100)^48 .......... - 100^50 - 50C_2(100)^48 + 50C_3(100)^47 ..........!!

!!B-A = 100xx(100)^49 + 50C_2(100)^48 .......... - 100^50 - 50C_2(100)^48 + 50C_3(100)^47 ..........!!

!!B-A = (100)^50 + 50C_2(100)^48 .......... - 100^50 - 50C_2(100)^48 + 50C_3(100)^47 ..........!!

!!B-A = 50C_2(100)^48 +50C_3(100)^47 .......... - 50C_2(100)^48 + 50C_3(100)^47 ..........!!

!!B-A = (50C_3(100)^47 + 50C_4(100)^46 ..........) + 50C_3(100)^47 - 50C_4(100)^46 ..........!!

Ignoring higher order terms we can see that !!B-A > 1!!, hence B is bigger.