shanmugamgsn wrote:
mangomangodolly wrote:
How to find last digit of below number.
\(101^102^103 * 102^103^104 * 108^109^110 * 109^110^111\)
Official answer is not provided of this question.
IMO unit digit must be 2..
As VIPS pointed out pls take care of ur post... Let it be legible....
Just edit this post...
Hi Shan, can you explain how did you come up with the answer 2 ?
Hi 2013gmat,
Sorry ans was not 2 as i calculated... it should be 4 if my evaluation is correct....
Here is my explanation, i'm ready to correct myself in case of mistakes...
\(a^b^c =\) a^(bc)
say
2^3^2 = \(2^6 = 64\)
I used this logic here...
101^102^103 = 101^10506 ====> unit digit ===> 1^6====> 1
102^103^104 = 102^10712 ====> unit digit ===> 2^2====> 4
108^109^110 = 108^11990 ====> unit digit ===> 8^0====> 1
109^110^111 = 109^12210 ====> unit digit ===> 9^0====> 1
1*4*1*1 = 4
correct me if im wrong
Shan, does the question say (a^m)^n OR a^(m)^n ? You have considered it as an former one but if it would have been the latter one.. then what wud be our approach ?