Latest TCS Aptitude Question SOLUTION: 2988^678 what is its last three digits? can anybody give its easy soln

write 678 in binary format its (1010100110)base2
the last 3 digits can be obtained by writing the last 3 digits of 988^678
now 988^1 = 988
988^2 mod 1000 = 144
988^4 mod 1000= 144* 144 mod 1000= 736
988^8 mod 1000 = 736 * 736 mod 1000= 696
988^16mod 1000 = 696*696 mod 1000=416
988^32 mod 1000= 416 *416 mod 1000 = 056
988^64 mod 1000 = 056 * 056 mod 1000= 136
988^128 mod 1000 = 136*136 mod 1000= 496
988^256 mod 1000= 496*496 mod 1000= 016
988^512 mod 1000= 016*016 mod 1000 = 256
now as we already have 678 in binary form we can write it as 512+128+32+4+2
988^(512+128+32+4+2) can be calculated as
(256*496*056*736*144)mod 1000 // using the mod 1000 values calculated above//
((256*496)mod 1000*(056*736)mod 1000*144)mod 1000
(976*216*144)mod 1000
= 504
this may not take a long time atleast with a calc

Point to be remembered : the last three digits of any multiplication comes by the multiplication of the last three digits of the numbers
for eg. 1121 * 1154 = 1293634
if we take the last three digits of the numbers which are :

121 * 154 = 18634

we are getting 634 in both of the results as the last three digits

Answer :

2988^678

we will consider only the last three digits that are 988

so now we will find 988^678

=> 988^(2*339)
we will find the square of 988 i.e. 976144
but will only consider the last three digits i.e. 144
we need to find now
=> 144^339
=> 144^(3*113)
we will find the cube of 144 i.e. 2985984
Considering only the last three digits, we now have to find

=> (984)^113
=> (984^112)*984

using the same concept

=> ((984)^(2*56))*984

=> (256^56)*984

=> (256^(2*28))*984

=> (536^28)*984

=> (536^(2*14))*984

=> (296^14)*984

=> (296^(2*7))*984

=> (616^7)*984

=> (616^6)*(616*984)

considering the last three digits of result of 616*984 i.e. 144

=> (616 ^ (2*3))*144

=> (456 ^ 3)*144

=> (456 ^ 2)*456*144

=> 936 * 456 * 144

=> 816*144

the last three digits are 504

it is easy but lengthy

(3000-12)^678
(12)^678-678*3000*(12)^677+_____________________
the last three digits are the last 3 digits of 12^678

first calculate 678/4( we get 2 as remainder).now compute 88^2 and we can get 744 as last three digits....so ans is 744.

(3000-12)^678
on expansion the terms contain 30000 figure will be atleast three 000 in last
hence the ans will be 12^678

i can give the solution upto 2 digits

first divide 678 by 4 and check cyclicity of 8..that will give 4 so the units place will be 4.

now for tens digit take 8(3rd one from lhs) and multiply it with 8(the power's unit digit ) that will give 64 .. so taking its unit place that will be 4..so the last 2 digit will be 44..but i don't have the idea of how to calculate the 3rd one..

ans cannot be 504 or any 144... unit digit of 8^8 is 6...so the unit digit of the ans must be 6.

988^2=87144;
so, Last 3 Digit = 144