Latest TCS Aptitude Question SOLUTION: 0, 1, 4, 15, 64, ____ 1) 225 2) 325 3) 125 4) None

answer is 325.
logic : 0*1+1,1*2+2,4*3+3,15*4+4,64*5+5,.....


number*1+1...result*2+2...result*3+3......

0,1,4,15,64.....
0*1+1=1
1*2+2=4
4*3+3=15
15*4+4=64
64*5+5=325

answer is 325

what is the remainder when 48^565 divided by 7? todays my tcs question

0*1+1=1
1*2+2=4
4*3+3=15
.
.
64*5+5=325

325..
no explanation

0^2=0
1^2=1
2^2=4
4^2-1 =15
8^2 =64
16^2-1 =255

0,1,4,15,64.....
0*1+1=1
1*2+2=4
4*3+3=15
15*4+4=64
64*5+5=325

answer is 325

Let n be the no. of objects.
By
V(k,n) = (n*k+1)*(V(k,n-1) + k^n)),
V(k,0) = 1.
n>=0
By changing K's value we get,
[k=0] 1, 1, 1, 1, 1, 1....
[k=1] 1, 4, 15, 64, 325, 1956...

Answer -> 2)325.

answer is 325.

64*5+5=325

2) 325
1*2+2=4
4*3+3=15
15*4+4=64
64*5+5=325

Answer shouls be 325

2>
0*1+1=1
1*2+2=4
4*3+3=15 ....And so on..64*5+5=325

325
0,0*1+1,1*2+2,4*3+3,15*4+4,64*5+5

0,1,4,15,64.....
0*1+1=1
1*2+2=4
4*3+3=15
15*4+4=64
64*5+5=325

answer is 325

1*2+2=4
4*3+3=15
15*4+4=64
64*5+5=325
so ans is 325

ans is 325.its like 1*2+2,4*3+3,15*4+4,64*5+5

0,1,4,15,64.....
0*1+1=1
1*2+2=4
4*3+3=15
15*4+4=64
64*5+5=325

answer is 325
Read more at http://www.m4maths.com/frequently-asked-placement-questions.php?ISSOLVED=&page=1&LPP=10&SOURCE=TCS&MYPUZZLE=&UID=&TOPIC=&SUB_TOPIC=#pO1Gvf4TZ7bD1Xbu.99

@ pervali yesaswi

Answer for u r question ....
48^565 / 7 = the number ( 48^565 )
Base is 48
Power is 565
Now, we can write power as 5*113 we can assume that ,
( 48^5^113) .
First step 48^5 = 48*48*48*48*48( multiply all the last digits of the numbers ) which results in (8*8*8*8*8)=(64*64*8) again multiply the last digits....(4*4*8) = 16*8 again last digit 6*8 finally u get 48 .in 48 last digit is 8 make a note of this ...
Now consider,

48^565=48^5^113=8^113
Logic= for every five times of 48 multiple we got last digit as 8. 113 has five multiple of about 22 and multiply of 48 for three times ( 113/5= 22 plus 3 as remainder).
For all the five multiples we can get the last digit as 8.

5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110

8,8,8,8,8,8,8,8,8,.......so on
Consider each five multiple as 8. we have 4 five multiples in the above multiple s of five and two are remained called 105,110.
Till 100= ( we have four multople of 5 multiples of 8 is 8) . hence 48 ^100= 8 is the last digit. And more 13 times we have five multiple is 2 and 3 remains .
48*48*48*48*48*48*48*48*48*48*48*48*48=[ total 13 times ]…
For first five times last digit is 8. And next five times also 8. And we have last three times now the final conclusion is

Last digit of 100 times is 8.
Last digit of five times is 8 and again 8 for next five digits. And finally 3 times 48.

==(8*8*8*8)*(8*8)*(48*48*48)
Multiply only last digits
==(8*8*8*8)*(8*8)*(8*8*8)

First five 8 are last digit is 8.
Now = 8*8*8*8*8= last digit is 8.


------- 8/7==remainder 1.

325 :: 4 ka 3 se multiply plus 3 add so we get 15 ! same process with 15 but with multiplier 4 and add 4 ....64*5 plus 5