Latest TCS Aptitude Question SOLUTION: find the last digit of the following expression:4+9^2+4^3+9^4+4^5+9^6+....+9^100

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find the last digit of the following expression:4+9^2+4^3+9^4+4^5+9^6+....+9^100

Asked In TCS Elangovan A (7 years ago)

Unsolved Read Solution (9)

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Pritam Dutta  (7 years ago)

so the given problem is......
4+9^2+4^3+9^4+4^5+9^6+....+9^100
now we can separate this series into two segments as follows....
( 4+4^3+4^5+....4^99)+(9^2+9^4+9^6+....+9^100)
first we will consider the fast segment....
4 last digit is 4
4^3 =64 last digit is 4
4^5=1324 last digit is 4
last digit of every component in this segment is 4
so the last digit of this segment of this segment will be the last digit of 4*50(as of there are 50 elements in the First segment(99+1/2))
that is 200
now we will consider the second segment....
9^2 last digit is 1
9^4 last digit is 1
9^6 last digit is 1
last digit of every component in this segment is 1
so the last digit of this segment of this segment will be the last digit of 1*50(as of there are 50 elements in the First segment(99+1/2))
that is 50
so the last digit of the given problem will be 200+50=250------->0
o the last digit of the given problem will be--->0(Ans)

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Steephen Joseph  (7 years ago)

The sum goes like 4+81+64+81*81+.........
That is the sum of units digit goes like 4+1+4+1+..........up to 100 terms.
We can consider (4+1+4+1) as a set of 4 numbers and the sum is 10. So 100/4 =25. That is 25 more sets are there and the answer therefore is 25*10= 250 (total sum). So last digit is 0 (Answer).

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RUPAM BANERJEE  (7 years ago)

last digit of 4^1 = 4 ; 4^3 = 4 ; 4^5= 4....
So, 99 = 1+(n-1)*2 n = 50 i.e. 4+4+4....50 times i.e. 4*50 = 200 (last digit 0)
Similarly, last digit of 9^2=1 ; 9^4 = 1....
So, 100 = 2+(n-1)*2 n=50 i.e. 9+9+9+....50 times i.e. 9*50 = 450 (last digit 0)
So, 200+450 = 650 i.e. last digit is '0'.

Is this solution Helpfull? Yes (2) | No (3)  

R Devadas Alva  (7 years ago)

Ans is 0 .

Is this solution Helpfull? Yes (1) | No   

gopal  (7 years ago)

Ans is 0 .

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MARIAMMAL@MANJU.B  (7 years ago)

use unit digit cycle for digit 4and 9 so u will get last digit as 0.

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dilpreet singh  (7 years ago)

There are two series here: series 1 and series 2
In the series 1:-
4+4^3+4^5+.....+4^99
last digit is 0
In the series 2:-
9^2+9^4+9^6+.....+9^100
last digit is also 0 here.

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Vijay  (5 years ago)

#include
int isPrime(int k)
{
int flag=0;
if(k==2)
{
flag = 0;
}

for(int j=2; j

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Vidya  (5 years ago)

4^1=4
4^2=16
4^3=64
4^4=256 i.e., the digit at units place can be either 4 or 6
similarly , for 9 the digit at units place can be either 9 or 1
so the above sequence for knowing last digit can be written as:
4+1+4+1+4+1...........4+1=50(4)+1(50)=200+50=250
hence the last digit is 0

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