TCS
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Numerical Ability
Number System
which of the following is power of 3?
a.2345
b.9875
c.6504
d.9833
Read Solution (Total 24)
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- option c is the right answer because 6+5+0+4=15 which is divisible by 3.
but in option a, 2+3+4+5=14 ,not divisible by 3
in option b, 9+8+7+5=29 ,not divisible by 3
in option d, 9+8+3+3+=23, not divisible by 3, - 8 years agoHelpfull: Yes(11) No(3)
- sum of the digits=6+5+0+4=15 which is divisible by 3.(opt c)
- 8 years agoHelpfull: Yes(7) No(3)
- power of 3 has 1,3,7,9 in their unit digit place hence ans:d
- 8 years agoHelpfull: Yes(7) No(0)
- 2^3=8
is 8 divisible by 3???
i tnk the approaches are wrong - 8 years agoHelpfull: Yes(3) No(2)
- Here power denotes perfect product or dividable number .. here 6504 only is dividable by 3
answer: 6504 ( C ) - 8 years agoHelpfull: Yes(1) No(0)
- sum of digits is divisible by 3 .so,option C is right
- 8 years agoHelpfull: Yes(1) No(0)
- none of the above
- 8 years agoHelpfull: Yes(1) No(1)
- none of the above ans is right 3^7=2187, 3^8=6561,3^9=19683.
- 8 years agoHelpfull: Yes(1) No(0)
- Option c is write bcz 6504 is divisible by 3 (or)
Sum of digits 6+5+0+4=15 which is divisible by3
Other options are not divisible by 3 - 8 years agoHelpfull: Yes(0) No(0)
- Ans: Option c-6504
Because 6504 is divisible by 3 - 8 years agoHelpfull: Yes(0) No(0)
- Since only option (c) 6504 adds upto 6+5+0+4= 15 is divisible by 3 therefore only 6504 is a power of 3.
- 8 years agoHelpfull: Yes(0) No(0)
- 2345/3 is not divisible by 3 similarly b and d also since option c is divisible by 3 so it is correct
- 8 years agoHelpfull: Yes(0) No(0)
- If any number is in power of 3, then it should also be divided by 3... if we see options only c.6504 is divisible by 3... soo answer is C.6504
- 8 years agoHelpfull: Yes(0) No(0)
- 6504 is a divisible by 3
- 8 years agoHelpfull: Yes(0) No(1)
- Power of 3's are always ends with 3, 9, 7, 1
- 8 years agoHelpfull: Yes(0) No(0)
- unit digit is 3 so answer is d
- 8 years agoHelpfull: Yes(0) No(0)
- Sorry guys you all are wrong
correct ans is D)9833.
we know that 3^1=3,3^2=9,3^3=27,3^4=81,3^5=243 and so on...
here see last digit 1,3,9,7 only it again repeated..
and in option there is only one that has last digit 3 - 8 years agoHelpfull: Yes(0) No(2)
- First check divisible to 3.
2+3+4+5=14 not divisible ,9+8+7+5=29 not divisible, 6+5+0+4=15 divisible, 9+8+3+3=23 not divisible
Here only one Number is divisible 6504 hence
Ans c) - 8 years agoHelpfull: Yes(0) No(0)
- in the options only 6504 is divisible by 3 with remider equal to ' 0'.
- 8 years agoHelpfull: Yes(0) No(0)
- if the number is power of 3 it should be divisible by 3 so option a is answer
- 8 years agoHelpfull: Yes(0) No(0)
- 6504
becoz sum of digits is divisible by 3 unlike others
- 8 years agoHelpfull: Yes(0) No(0)
- 3,9,27,81 order is
3,9,7,1 so ans=d - 7 years agoHelpfull: Yes(0) No(0)
- last digit of 3^1=3, 3^2=9, 3^3= 7, 3^4= 1, 3^5= 3, 3^6=9 clearly we can see that we can never get 5 r 4 in the last digit with the power f 3 so d is the right answer
- 6 years agoHelpfull: Yes(0) No(0)
- C 6+5+0+4=15/3
- 6 years agoHelpfull: Yes(0) No(0)
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