Latest TCS Aptitude Question SOLUTION: last three digit of 2988^678

(2988)^687= ((2^2)*(3^2)*83)^(6*113)

= ((2^12)*(3^18)*(83^6))^113
= (2^1356)*(3^2034)*(..69^113)

now powers of 69 repeat itself after 69^10

so (..6)*(..9)*(..09)=486

This is a tough question.
Last three digits of an expression is nothing but the remainder when we divide the given expression by 1000.
So 29886781000=(−12)6781000=126781000
We write 1000 as 8 ×125. So we separately find the remainders by dividing the given number by 8 and 125.
Given expression is clearly divisible by 8. Also
Euler totient theorem says [aϕ(N)N]Remainder=1
Here ϕ(N) is the number of co primes of a number N. In this case ϕ(125) = 100.
So (12100)6.1278125=16.1278125
Now 1278=478.378=(1+15)39.(10−1)39
(1+15)39.(−1)39(1−10)39
- (1+39C1.15+39C2.152.... ). (1+39C1.10+39C2.102−39C3.103....
If you observe clearly, 153 and 103 are 15 multiples. So we can omit the remaining terms.
-(1+39.15 + 39.19.152 ).(1-39.10 + 39.19.102)
By writing 39 as 40-1 we can simplify this equation fast.
-(1+(40-1).15 + (40-1).19.152).(1-390 + (40-1).19.100
-(1+100 - 15 - 19.152 ).(1-140 - 19.100)
-(101 - 15 - (20-1).152).(1-140+100)
-(101-15+100).(-39)
-(61).39
-2379 = 4
So the remainder is 4.
The original expression is in the format of N = 8K = 125 L + 4
The minimum number satisfies this condition is L = 4.
So 125 (4) + 4 = 504
So answer is 504