Latest TCS Aptitude Question SOLUTION: The last digit of the expression 4+92+43+94+45+96+....+499+9100 is.

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(#M40023681) TCS Numerical Ability question Keep an EYE

The last digit of the expression 4+92+43+94+45+96+....+499+9100 is.

Asked In TCS priyanka (10 years ago)

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keerthana  (10 years ago)

3 :
43+45+.........499 are in a.p. find sum of digits using n/2(2a+(n-1)d)
last digit of this sum is 9. similarly 92+94+...........9100.,last digit of this sum is 0.. soo 9+4+0 = 13. 3 is last digit.

Is this solution Helpfull? Yes (29) | No (4)  

keerthana  (10 years ago)

7 :
43+45+.........499 are in a.p. find sum of digits using n/2(2a+(n-1)d)
last digit of this sum is 9. similarly 92+94+...........9100.,last digit of this sum is 4.. soo 9+4+4 = 17. 7 is last digit.

Is this solution Helpfull? Yes (10) | No (8)  

Karthika  (10 years ago)

Yeah Keerthana! 3 is the right answer!

Is this solution Helpfull? Yes (6) | No (3)  

Karthika  (10 years ago)

In the sum of 92+94+...9100 , the last digit of this sum is "0" ryt??

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Nirmalya Basu  (9 years ago)

Two different AP series exist here :

1) 43+45+47+....+499
2)92+94+96+....+9100

499=43+(n-1)2
=>n=229
Sum=229*(43+499)/2=Last digit will be 9

9100=92+(n-1)2
=>n=4459
Sum=4459*(92+9100)/2=Last digit will be 4

So sum of last digits is 4+9+4=17
Hence last digit is 7

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Shatarupa Chaudhuri  (9 years ago)

43+45+...+499 are in AP ...after finding out the no of terms using tn=a+(n-1)d, we get the sum of the AP series....the last digit of which comes to 9...by using the same process for 92+94+....+9100, the last digit for it comes to 0..thus the last digit of the whole series becomes 9+0=9. Thus answer is 9.

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sivaram  (7 years ago)

92+94+96+...then 43+45+47+....the series end with a even number. 501

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RITU YADAV  (7 years ago)

43+45+………499 are in a.p.
find sum of digit using n/2(2a+(n-1)d)
last digit of this sum is 9.
Similarly 92+94+………..9100.
Last digit of this sum is 0.
So 9+4+0 = 13.
3 is last digit.

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