exponent question from Kaplan CD-ROM

damorris wrote:Hi,

hit the proverbial brick wall with this practice question from the Kaplan CD-ROM:

What is the units digit of (9)^5(13)^3(7)^3?

(Can't remember the answer choices)

Does anyone know how to arrive at an answer without a lot of number crunching?

Just consider only unit digit of all terms
ie unit digit of 9^5 --->9 (all powers of 9 has a cyclicity of 1 & 9 in unit digit ie 9^1=9, 9^2 = 81, 9^3=729)

similarly 13^3 ---->7

7^3 ---->3

so unit digit of (9)^5(13)^3(7)^3 = unit digit of 9*7*3= 9 8)

Just see the cyclicity of units digit when 9,3(in 13) and 7 are mulitplied

For 9(9*9*9*9.....) - > 9,1,9,1...

For 3(3*3*3*3...)-> 3,9,7,1,3,9,7,1....

For 7 -> 7,9,3,1,7,9,3,1

We are interested in the bold numbers above based on the powers to which each of the numbers 9,3,7 are raised

So 9*7*3 = 189

Answer should be 9

Thanks for that. It certainly helps with (9)^5

. . . but doesn´t 7^3 ---->3 ? (343)

Therefore the answer is 9*7*3 = 9

Right it's 9

For 9(9*9*9*9.....) - > 9,1,9,1,9,1

I higlighted the 3rd in bold instead of 5th in my previous post (it doesnt matter since 9,1 alternates so if 9 is in 3 9 will be there in the 5th,7th,9th etc...)

I corrected it anyway!