Latest TCS Aptitude Question SOLUTION: Find the sum of all the four digit numbers that can be formed using the digits 2,3,5,9. No digit is to be repeated.

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(#M40023730) TCS Logical Reasoning question Keep an EYE

Find the sum of all the four digit numbers that can be formed using the digits 2,3,5,9. No digit is to be repeated.

Asked In TCS Prakash Singh (10 years ago)

Solved chennadeepthi Read Solution (17)

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Read Solution (17) :

priya  (10 years ago)

formula:
sum of digits*(n-1)!*(111....number composed of n digits)
soln:
19*3!*1111=126654

Is this solution Helpfull? Yes (65) | No (2)  

chennadeepthi  (10 years ago)

Be careful that your number has four distinct digits, otherwise the counting gets more confusing.

There are 4!=24 such numbers. Add them up, without carrying first.

In each column, you will have each of the four digits appear six times. So each column, without carries, will add up to 6(a+b+c+d). So the sum is

6(a+b+c+d)(1+10+100+1000)=
6(a+b+c+d)*1111=
6666(a+b+c+d).
=6666(2+3+5+9)=6666(19)=126654

Is this solution Helpfull? Yes (33) | No (4)  

Anusha  (10 years ago)

answer:126654
explanation:(n-1)!(sum of all the digits)(1111...n times)
here n=4.so,(4-1)!(2+3+5+9)(1111)

Is this solution Helpfull? Yes (7) | No (1)  

k.m  (10 years ago)

answer is 226654

Is this solution Helpfull? Yes (4) | No (8)  

rajat shah  (10 years ago)

@PRIYA
what is the solution if digit is to be repeated....?
and what is formula for it...??

Is this solution Helpfull? Yes (4) | No   

Aman Agarwal  (10 years ago)

There are 4! = 24 numbers that can be formed. Now in each column we have each digit 6 times. So keeping that in mind,
For unit's place sum is: (6*2)+(6*3)+(6*5)+(6*9)=114 --> 4 at units place and 11 in carry
For ten's place sum is: (6*2)+(6*3)+(6*5)+(6*9)+11=125 -->5 at ten's place and 12 in carry
For hundredth place sum is: (6*2)+(6*3)+(6*5)+(6*9)+12=126 -->6 at hundredth place and 12 in carry
For thousand place sum is: (6*2)+(6*3)+(6*5)+(6*9)+12=126 -->6 at thousandth place and 12 in carry
So combining the result we get: 126654

Is this solution Helpfull? Yes (3) | No   

Prakash Singh  (10 years ago)

give the logic

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

Divya  (10 years ago)

answer is [6*1000*(2+3+5+9)]+[6*100*(2+3+5+9)]+[6*10(2+3+5+9)]+(6*1*(2+3+5+9)]=126654

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

prarit jain  (10 years ago)

sum of digits is asked so it sould be 19 * no. of ways ie 4! so ans is 19*24= 456

Is this solution Helpfull? Yes (2) | No (6)  

sithick mohammed  (9 years ago)

incase they ask with repetition of numbers..wat is the solutio

Is this solution Helpfull? Yes (2) | No   

ch sathwik  (10 years ago)

i think answer is 126654

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

juttigakrantija  (10 years ago)

we can arrange 2,3,5,9 in 16 ways and sum of 2,3,5,9 is equal to 19
therefore 16*19=304

Is this solution Helpfull? Yes | No (10)  

palakurthy tanuja  (10 years ago)

in each column digit will be repeated six times as per solution..how 6 times?

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LAXMI DURAI  (10 years ago)

it will be 456 only....bcz 1st the digit starts with 2 and arrangements in this is may be 2359 ,2395,2539,2593,2935,2953 here 6 possibilities and each sum is 19 and 19*4= 114 and the sum became 114*6 for the num starting with 2,3,5,9 and it will be 456

Is this solution Helpfull? Yes | No (2)  

Tulasi Ram  (7 years ago)

2+3+5+9=19

Is this solution Helpfull? Yes | No (1)  

Mahalakshmi. V  (7 years ago)

given is 2,3,5,9
then 4!=24 and which is divided by 4 will give 6
6(1000+100+10+1)*(2+3+5+9)=6666(2+3+5+9)
then the answer is 126654

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Namratha  (6 years ago)

can any one tell me how to solve this if the digits can be repeated?

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