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Numerical Ability
What is the unit place of the sum
(12)^41+(66)^66+(25)^15+(51)^61+4321
a)1
b)3
c)8
d)6
Read Solution (Total 13)
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- anser is 5 all options are wrong
- 9 years agoHelpfull: Yes(9) No(0)
- 2+6+5+1+1=15
so unit place will be 5 - 9 years agoHelpfull: Yes(7) No(0)
- 12^41=last digit=2
66^66= last digit=6
25^15= last digit=5
51^61= last digit=1
4321 = last digit=1
so last digit=2+6+5+1+1=15
so last digit=5
- 9 years agoHelpfull: Yes(5) No(0)
- I think the answer is 6
- 9 years agoHelpfull: Yes(3) No(5)
- plz explain the and...
- 9 years agoHelpfull: Yes(1) No(0)
- ans will be 6
take first term 12^41,concentrate on 2 (unit digit) and 41
the cycle of 2 is 4 (2,4,8,16,32)
so divide 41 by 4 ,remainder is 1 so unit digit of 12^41=2^1=2
similarly cycle of 6 is 1 ,remainder is 0 so unit digit of 66^66=6^0=1
same for 25^15 unit digit=1
51^61=1
so adding 2+1+1+1+1=6 - 9 years agoHelpfull: Yes(1) No(10)
- Answer is 5.......
- 9 years agoHelpfull: Yes(1) No(1)
- Ans; 5
- 9 years agoHelpfull: Yes(1) No(0)
- The answer is 3 (b)
- 9 years agoHelpfull: Yes(0) No(2)
- how? answer is 3,@@@@@@KROVVIDI ANANTH DUTT
can you please explain me. - 9 years agoHelpfull: Yes(0) No(0)
- Ans is 3(b)
- 9 years agoHelpfull: Yes(0) No(0)
- Answer will be 5
- 9 years agoHelpfull: Yes(0) No(0)
- It's a dummy question. Answer is 5.
- 9 years agoHelpfull: Yes(0) No(0)
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