Latest TCS Aptitude Question SOLUTION: when 101102103104............150 is divided by 9 what is the remainder?

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when 101102103104............150 is divided by 9 what is the remainder?

Asked In TCS thilaga (9 years ago)

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karthiga  (9 years ago)

1(01) 1(02) 1(03)................1(50)
sum of the digit = 1325
1325/9 remainder is 2.
(or)
101 102 103 104..... 150
sum of the digit = 150*151/2 - 50*51/2;
= 11325 - 5050
= 6275/9
remainder is 2

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Suraj Kumar Shresth  (5 years ago)

sum of the n digits = n/2[a+l].
here n=50 , a=101 and l=150
so , 50/2 [101+150]
25 * 251 = 6275.
and when 6275/9 , rem=2 .
hence answer 2.

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Vivek shaw  (9 years ago)

when 101/9---remainder will be 2.
when 101102 is divided by 9 remainder will be 4..
hence it will increment by 2 every time.

150/3=50.
that is 50*2=100

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sundaresh.k  (5 years ago)

101 102 103 104..... 150
sum of the digit = 150*151/2 - 100*101/2;
= 11325 - 5050
= 6275/9
remainder is 2

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Shashank Sahu  (4 years ago)

The divisibility rule for 9 is sum of the digits is to be divisible by 9. So
We calculate separately, sum of the digits in hundreds place, tenths place, and units place.
Sum of the digits in hundreds place: 1 x 50 = 50
Sum of the digits in tenths place : 0 x 9 + 1 x 10 + 2 x 10 + 3 x 10 + 4 x 10 + 5 x 1 = 105
Sum of the digits in units place : (1 + 2 + 3 + ...+ 9) x 5 = 225
So total = 380
So remainder = 380 / 9 = 2

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