JMO, indices, number

In the sum 23¹⁵ + 19¹⁷ + 27³⁵, find the digit in the unit place.

JMO STD V and STD VI

Answer:

23¹⁵ + 19¹⁷ + 27³⁵,

23¹⁵ = {(23)⁵}³

3⁵ has 3 in its units place.

So, {(23)⁵}³ will have 3 x 3 x 3 = 7 in the units place

19¹⁷ = {(19)³}⁵ x 19 x 19

(19)³ has 9 in the units place and so, {(19)³}⁵ = (....9)⁵ = (..9)³ (.9)² = 9 in the units place

So, 19¹⁷ =  9 x 9 x 9 = 9 in the units place.

27³⁵ = {(27)⁵}⁷.

(27)⁵ = 7 in the units place. And so (.....7)⁷ = (....7)⁵ x (...7) x (...7)

          = 3 in the unit place

So, sum of digits of units place = 3 + 9 + 7 = 9 in the units place