Question

Find the last 3 digits of
25^63 × 63^25

Answer 1

Answer is 375

The first thing that you have to realize is the interesting property of 5: % raise to any power, the last digit will always be 5; 25 raise to any power, the last 2 digits will always be 25; 625 raise to any power, last 3 digits will always be 625. Therefore, 25^63 can be written as (625^31)*25, which will give the last 3 digits as 625*25=15625. Hence, the last 3 digits of the left side will be 625.

Now, coming to 63^25. This can be written as (60+3)^25. Using the binomial theorem, we get 25C0 60^25 3^0  + 25C1 60^24 3^1  + ....  +  25C23 60^2  3^23   +  25C24 60^1  3^24   +25C25 60^0  3^25 . Now, if you carefully observe, all the expansion terms containing power of 60 as greater then equal to 3 will have last 3 digits as 0. Thus if we multiply these with 625, we will still get last 3 digits as 000. Thus, for 63^25, we need to focus on 25C23 60^2  3^23   +  25C24 60^1  3^24   +25C25 60^0  3^25  to get the last 3 digits. Now, 25C23 60^2  3^23  = 25*24/2  *3600 *3^23.  This terms also gives last 3 digits as 000. 

Now, we have 2 terms left in the expansion: 25C24 60^1  3^24   +25C25 60^0  3^25 .  Taking 25C24 60^1  3^24= 25*60*3^24=1500*3^24. Thus, last 2 digits of this equation will definitely be 00. We need to know the 3rd digit which will be 5* units digit of 3^24. Now, 3^4 is 81, thus, units digit of 3^24 will also be 1. Hence, the last 3 digits of 25C24 60^1 will be 500. 

Our current equation looks like: 625*(500+ 3^25). Finally coming to the last expansion: 25C25 60^0  3^25=3^25. This looks troublesome. Our current equation looks like: 625*(500+ 3^25). I am thinking of using another interesting property of 625: if we multiply 625 by 8, the last 3 digits will be 000. Now, using binomial theorem on 3^25 as (2+1)^25, we get   25C0 2^25   + 25C1 2^24   + ....  +  25C23 2^2    +  25C24 2^1   +     25C25 2^0  . Since, 625 * 8 gives last 3 digits as 0, we are only left with terms which have power of 2 less than equal to 2. Thus, or expansion terms now are  25C23 2^2    +  25C24 2^1   +     25C25 2^0 which is 25*24/2*4+25*2+1=1200+50+1=1251. Hence the last 3 digits for this expansion term are 251. 

Hence, the entire thing can be boiled down to 625* (500+251)=625*751=469375. Hence, the last 3 digits are 375