Latest TCS Aptitude Question SOLUTION: 3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13 a. 0 b. 12 c. 3 d. None of these

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(#M40041344) TCS Numerical Ability question Keep an EYE

3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13

a. 0
b. 12
c. 3
d. None of these

Asked In TCS Sanu Panji (9 years ago)

Unsolved Read Solution (5)

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harshit gupta  (9 years ago)

ANS is 0
Lets see this cycle
3^0%13=1
3^1%13=3
3^2%13=9
3^3%13=1
3^4%13=3
3^5%13=9
..... and so on, as it repeat after 3 power, so remainder remain same for more powers
Now for adding 1st three remainder =1+3+9=13/13=0
after every three elements reaminder is 0
201/3=0 thus overall remainder is 0

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Sanu Panji  (9 years ago)

Ans is 0

3^0+3^1+...210 terms is geometric progression with 201 terms the sum of which is calculated using formulae to be 3^201 - 1/2

Now 3^201 is nothing but 3^20 X3^20..ten times X 3 and 3^20 has unit digit 1 so obviously the unit digit of while expression is 3. Thus 3^210 - 1 has unit digit 2 which when divided by 2 will give no remainder. Hence remainder is 0.

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deepak kumar  (9 years ago)

series =(1+3+(3^2)+(3^3)+(3^4)+(3^5)......)/13
({1+3+(3^2)}+(3^3){1+(3^1)+(3^2)}......)/13
{13+(3^3)13+......}/13
{1+3^3+.....}
thus no remainder

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sona sharma  (9 years ago)

the answer is zero
3^0=1 and 1%13=1
3^1=3 and 3%13=3
3^2=9 and 9%13=9
and 1+3+9=13 and 13%13=0
so the result is zero

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Janani  (9 years ago)

last term :
3^201 mod 13 => 201 mod 12 =9
3^9 mod 13 =1

sum=(3^201-1)/2=1-1/2=0

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