Latest TCS Aptitude Question SOLUTION: 123456789101112.... . 4465 . find the remainder when divided by 45?

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(#M40029970) TCS question Keep an EYE

123456789101112.... . 4465 . find the remainder when divided by 45?

Asked In TCS pre (10 years ago)

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RAKESH  (10 years ago)

123456789101112....4465
=123456789101112...4400+65
sum of digits of 123456789101112...4400=1+2+3+...+44=(44*45)/2=22*45
clearly 123456789101112...4400 is divisible by 9 & 5, so divisible by 45
hence when 65 is divided by 45,20 will be remainder
ANS=20

Is this solution Helpfull? Yes (23) | No   

sandeep gantait  (10 years ago)

in the given question first calculate the sum upto 44 terms by the formula n(n+1)/2
which comes:44(45)/2=(990+65)%9*5=Ans:2

Is this solution Helpfull? Yes (8) | No (15)  

vankayala krishna kanth  (10 years ago)

ans=0
given num is multiply 0f 45

Is this solution Helpfull? Yes (7) | No (11)  

harish budhwani  (10 years ago)

First calculate sum upto 465 terms as the last term is 465 using n(n+1)/2.
465(465+1)/2=108345.
Divide 108345 by 45 which gives remainder 30.

Is this solution Helpfull? Yes (6) | No (5)  

Pamela Das  (10 years ago)

ANS WILL BE 20

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

RAKESH  (10 years ago)

last term isn't clear
4465 or 464 465 or 65

Is this solution Helpfull? Yes (4) | No   

rajesh ray  (10 years ago)

Divisibility for 45 is, the given number should be divisible by 9 as well as with 5.
Let the given number is N.
N has unit digit 5 so it is divisible by 5. Now the divisibility for 9 is digit sum of N should be divisible by 9.
Digit sum of N = (1+2+3+...9) + 1 x 10 + (1+2+3+.....9) + 2 x 10 + (1+2+3....9) + 3 x 10 + (1+2+3.....9) + 4 x 5 + (1 + 2 + 3 + 4) + 6 + 5
= 45 + 10 + 45 + 20 + 45 + 30 + 45 + 20 + 10 + 11 = 2 (or 1 + 2 + 3+....44 + 65 = 44×452+65=990+65=1055=2
So N = 5K = 9L + 2
For L = 2 the given number gets satisfied. So the least number satisfies the condition = 9(2) + 2 = 20

Remainder = 20

Is this solution Helpfull? Yes (3) | No (1)  

Raushan shukla  (10 years ago)

20 wl b remainder. confrm ans

Is this solution Helpfull? Yes (2) | No (1)  

dipu  (10 years ago)

Divisibility for 45 is, the given number should be divisible by 9 as well as with 5.
Let the given number is N.
N has unit digit 5 so it is divisible by 5. Now the divisibility for 9 is digit sum of N should be divisible by 9.
Digit sum of N = (1+2+3+...9) + 1 x 10 + (1+2+3+.....9) + 2 x 10 + (1+2+3....9) + 3 x 10 + (1+2+3.....9) + 4 x 5 + (1 + 2 + 3 + 4) + 6 + 5
= 45 + 10 + 45 + 20 + 45 + 30 + 45 + 20 + 10 + 11 = 2 (or 1 + 2 + 3+....44 + 65 = 44×452+65=990+65=1055=2
So N = 5K = 9L + 2
For L = 2 the given number gets satisfied. So the least number satisfies the condition = 9(2) + 2 = 20

Remainder = 20

Is this solution Helpfull? Yes (2) | No (1)  

Pratyusha Bhadra  (10 years ago)

given nmbr is a multiple of 3...not of 9 @vankayala....

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SAPTARSHI MONDAL  (10 years ago)

465*466/2 =
remainder 2

Is this solution Helpfull? Yes | No (1)  

kirti  (10 years ago)

24C7*2*17!*18

Is this solution Helpfull? Yes | No (2)  

Sumit Chatterjee  (10 years ago)

First calculate the sum of the series using n2[2a+(n-1)d]
i.e 652[2+(65-1)] = 2145,now you can divide dis by 45 ul get 30 as remainder

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