Latest TCS Aptitude Question SOLUTION: what is the sum of all the 4 digit numbers that can be formed using all of the digits 2,3,5 and 7?

• Placement Papers
• /
• TCS

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40021887) TCS question Keep an EYE

what is the sum of all the 4 digit numbers that can be formed using all of the digits 2,3,5 and 7?

Asked In TCS reddy siri (10 years ago)

Solved Rehman Read Solution (8)

Is this Puzzle helpful?   (10)   (0) Submit Your Solution

Read Solution (8) :

Rehman  (10 years ago)

Actually there is the direct formula for this kind of problems.
1. Sum of all the numbers which can be formed by using the n digits without repetition is: (n-1)!*(sum of the digits)*(111…..n times).

2. Sum of all the numbers which can be formed by using the n digits (repetition being allowed) is: n^{n-1}*(sum of the digits)*(111…..n times).
so as per question:(4-1)!*(2+3+5+7)*1111===113322

Is this solution Helpfull? Yes (69) | No (2)  

hemanth kumar  (10 years ago)

113322
The value of 2 in once place is 2,in tens place is 20,in hundreds place is 200,in thousands plce is 2000,add all you get 2222,in the same way for 3,5 an 7 dat is 3333,5555 and 7777 respectively,now add all we get 18887,now here comes the problem fix the thousands digit and the number of ways the remaining can be arranged is 3! ways dat is 6,so 18887*6 is 113322

Is this solution Helpfull? Yes (6) | No   

Rudra  (10 years ago)

@Rehman Brilliant Man!!!!!

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

rahul singh  (10 years ago)

_ _ _ _ the 4 digit no. will be formed in 4*3*2*1 =24 ways so the digits 2,3,5,7 will be repeated 6 times each hence 2+3+5+7 = 17 and 17*6=102
11 11 10 10
x x x x
x x x x
___________
113322

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

Satyaki Kundu  (10 years ago)

18887

solution: (sum of all digits) * P (r-1, n-1) * (1111....r times)

where r=4 (r digit number)
and n=4 (2,3,5,7)

(2+3+5+7)*P(3,3)*(1111) = 18887

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

reddy siri  (10 years ago)

i know that the correct ans is 113322 but dnt know how to solve

Is this solution Helpfull? Yes (4) | No (3)  

rekha  (10 years ago)

there are 4^4 different number . each digt comes in the number is 4^4/4=4^3=64
unit digits takes 2,3,5,7 as 64 times and similar 10,100 and 1000.
sum=64*1000(2+3+5+7)+64*100(2+3+5+7)+64*10(2+3+5+7)+64*2+3+5+7)=1208268(i take it as repeatable);

Is this solution Helpfull? Yes (2) | No   

VIKAS SINGH  (10 years ago)

at every place all digits will occur six times so at unit place(2+3+5+7)*6=0 and carry 42 now at tens place the sum will be constant 420+42(carry)=2 nd carry 46 now at hundred place 420+46=6 and carry 46 . at thousend place 420+46=466.
so sum will be 466620.

Is this solution Helpfull? Yes (1) | No (8)  

Please Login to Read Solution.