Latest TCS Aptitude Question SOLUTION: Question : Given sequence 91011....33 .what is the remainder when divided by 9? a)2 b)0 c)3 d)7

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Question : Given sequence 91011....33 .what is the remainder when divided by 9?
a)2
b)0
c)3
d)7

Asked In TCS Amit Bansal (8 years ago)

Unsolved Read Solution (11)

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sonu sinha  (8 years ago)

c)3
(9+10+11+12.......33)-(1+2+3...8)
so here
n(n+1)/2-n(n+1)/2
(33x34)/2-(8x9)/2
33x17-4x9
561-36=525
now 525/9=3

Is this solution Helpfull? Yes (8) | No (2)  

Ashwin  (8 years ago)

It will be,
10's place=> 1*10 + 2*10 + 3*4 = 42
Unit's place=> 9+2(0+1+2+3+4+5+6+7+8+9)+(1+2+3) = 105

So,105+42=147
147/9=rem(3)

Is this solution Helpfull? Yes (5) | No (1)  

Sachin  (8 years ago)

S@ch!n
first term = 9
Last term = 33
total number of term = 25
sum = n(a+L)/2
sum = 25(9+33)/2 = 42*25/2 = 525
remainder = 525 / 9 = 3
Ans = 3

Is this solution Helpfull? Yes (5) | No (1)  

Angad  (8 years ago)

sum of total no of 1 at 10's place =10*1=10
sum of total no of 2 at 10's place =10*2=20
sum of total no of 3 at 10's place =3*4=12
sum of numbers at unit place=45*2=90
total sum =90+42+9+6=147
147/9 rem=3

Is this solution Helpfull? Yes (4) | No   

Ashwin  (8 years ago)

Add 9+10+11+....+33. You'll get 525. Divide it by 9, the remainder is 3.

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

manish yadav  (8 years ago)

Ans :- 0

sum of unit digit => 9 + unit digit of (10 to 30) + unit digit of (31,31,33 )
= 9 + 2 *(1+2+3+......+9)+1+2+3
= 9+ 90+6
=105

sum of ten's place => ten's place from (10 to 33)
= 10*1 (from 10 to 19)+ 10*2 (from 20 to 29) +4*3 (for 30,31,32,33)
=10+20+12
=42
total sum of unit and ten's digit = 105+42= 147
divided by 9 => 147 %9 = 3 will be remainder

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

Priyesh jain  (8 years ago)

sequence will be
9101112131415161718192021222324252627282930313233
divisibility rule of 9 is sum of all digits should be multiple of 9
sum of digits = 135%9==0;

Is this solution Helpfull? Yes (1) | No (6)  

Aparna kumari  (8 years ago)

33=a+(n-1)d
33=9+n-1
33-8=n
n=25
sum=25(9+33)/2
25*42/2
25*12=525
525/9=3

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Amit Bansal  (8 years ago)

Answer b) 0
sum of unit digit= (1+2+....9)*2 +1+2+3+9
=>sum=45*2+15
sum=105
sum of ten's place=10*1 +10*2+ 3*3
10+20+9
=39
=> total sum=39+105=144
which is completely divisible by 9

Is this solution Helpfull? Yes | No (7)  

shanoor  (8 years ago)

Arithmetic progression = 9+10+11+12.....33

where a=9, l=33,

sum of arithmetic progression = n/2(a+l) => 24/2(9+33) =>396

so, 396/9 gives the zero reminder.... Ans= 0

Is this solution Helpfull? Yes | No (6)  

prachi  (8 years ago)

Here, first term,a =9 and last term(l) is 33, n= 25(from 9 to 33).
Now, Sn= n/2[a+l]
Sn=23/2[9+33]
=525
Now, divide by 9, we get
525/9= 3(as Remainder).

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