GATE ECE 2014 Set 4 | GA | Question: 10

Question

A five digit number is formed using the digits $1,3,5,7$ and $9$ without repeating any of them. What is the sum of all such possible five digit numbers?

• A. $6666660$
• B. $6666600$
• C. $6666666$
• D. $6666606$

Comments

Answer Here!   

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number of 5 digits numbers is 120 and 120 divides only option B

120 should divide sum of all numbers because sum of all 120 numbers is asked.

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THNKS

Answer 1

(B)

Consider the digits $1,2,3$.

The possible numbers are $\{123,132,213,231,312,321\}$, count  = $3! = 6$.

• Consider the cases where the digit $3$ is at the unit position. The number of such numbers (fix $3$ at unit, permute the rest) will be $(3-1)! = 2!$. In all such cases, the $3$ digit will contribute a $3$ to the final sum. Total = $2! \times 3$
• Consider the cases where the digit $3$ is at the decimal position. The number of such numbers  (fix $3$ at decimal, permute the rest) will be $(3-1)! = 2!$. In all such cases, the $3$ digit will contribute a $30$ to the final sum. Total = $2! \times 30$
• Consider the cases where the digit $3$ is at the hundreds position. The number of such numbers  (fix $3$ at hundreds, permute the rest) will be $(3-1)! = 2!$. In all such cases, the $3$ digit will contribute a $300$ to the final sum. Total = $2! \times 300$

In total, the digit $3$ contributes $2! \times (3+30+300) = 2! \times 333$ to the final sum.

The same happens for all other digits.

Hence, the net sum (for the original question) will be:

$(5-1)! \times (11111 + 33333 + 55555 + 77777 + 99999)$

Note: $(5-1)!$ = the number of permutations after fixing $1$ digit. $11111 \ldots$ because the digit $1$ contributes a $1, 10, 100, 1000, 10000$ to the final sum.

Required Answer = $6666600$.

Hence, option B is correct.

Comments

For more explanation

(for non zero digits)https://www.youtube.com/watch?v=PQ5OySVcvSQ

(digits which include zero)https://www.youtube.com/watch?v=-3UDo6qEi2w

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Very good approach

Answer 2

There are $5!$ numbers possible without repetition of digits and each of the five digits repeats $4!$ times at left most position and similarly for all other digit positions this repetition is $4!$. So, we can get the sum of numbers as

$ 4! \times (1+3+5+7+9) \times 10^4 $
$+ 4! \times (1+3+5+7+9) \times 10^3 $
$+ 4! \times (1+3+5+7+9) \times 10^2 $
$+ 4! \times (1+3+5+7+9) \times 10 $
$+ 4!(1+3+5+7+9) $
$= 6000000 + 600000 + 60000 + 6000 + 600 $
$= 6666600.$

Comments

This formula can further be simplified as:

(|n|-1)! *(11... m times)*(sum of the digits in set n)

where

m is the no of digits in the resultant number

n is the set of digits (in this case {1,3,5,7,9})

|n| is cardinality of n

Answer 3

Number of combinations = $5!$

There are 120 combinations of 5 digit number consisting of 1,3,5,7,9

So the most obvious thing we can extract from this information is that every digit $(1,3,5,7,9)$ will be occurring 120 times and also that every digit will be occurring $\frac{120}{5} = 24$ times at each place value.

sum of one's position while adding = $(24 * 1) + (24 * 3) + (24 * 5) + (24 * 7) + (24 * 9) = 600 $

Carrying 60 to tens place we got ones digit of sum which is $0$

now adding all digits at tens place $600+60 = 660$

Carrying 66 to hundreds place we got tens place digit which is $0$

now adding all digits at hundreds place $600+66 = 666$

Carrying 66 to hundreds place we got tens place digit which is $6$

we already ommited all the options and are left with  option B with just last 3 digits of the sum but still adding and carrying will give $6666600$

Comments

just amazing!!

Answer 4

Using digits $1\mid 2\mid 3\mid \dots \mid n$

Sum of all possible $n$-digit number $ = \dfrac{n!}{n}(1+2+3+\dots + n)(10^{0} + 10^{2} + 10^{3} + \dots + 10^{n-1})$

Here,digits are $1\mid 3\mid 5\mid 7\mid 9$

Sum of all possible $5$-digit number $ = \dfrac{5!}{5}(\underbrace{1+3+5+7+9}_{\dfrac{5}{2}(1+9) = 25})(\underbrace{10^{0} + 10^{2} + 10^{3} +10^{4}}_{\dfrac{10^{5}-1}{9} = 11111})$

$ \implies \dfrac{120}{5}\times 25 \times 11111 = 600\times 11111 = 6666600 $

So, the correct answer is $(B).$

Comments

@Lakshman Patel RJIT sir can you plz explain this silly doubt 

M Not able to visualize why divided by 5?

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We need to form a $5$-digit number.

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@Lakshman Bhaiya according to ur formula common ratio in second term would be inconistent as 1st term is 1 and second is 100 ... u have missed 10^1 in that ..plz add that

Answer 5

So since the repetitions are not allowed, there will be total 120 permutations. Now if we look carefully at place values  , every digit is repeated 24 times in all the entire permutation. Hence for UNITS place there are 24  nines, 24 sevens, 24 fives, 24 threes, 24 ones.

So unit place sum = 24(9) + 24(7) + 24(5) + 24(3) + 24(1)= 600  ------→ 0

taking carry of 60 to tens place,  tens place sum=  24(9) + 24(7) + 24(5) + 24(3) + 24(1) + 60= 660   → 0

taking carry of 66 to hundereth place, hunderedth place sum=  24(9) + 24(7) + 24(5) + 24(3) + 24(1) + 66= 666  → 6

taking carry of 66 to thousandth place, thousandth place sum=  24(9) + 24(7) + 24(5) + 24(3) + 24(1) + 66= 666  -→ 6

taking carry of 66 to ten thousandth place, ten thousandth place sum=  24(9) + 24(7) + 24(5) + 24(3) + 24(1) + 66= 666---→ 6

taking carry of 66 to lakh place, lakh place sum=  0 + 66= 66

hence combining all the place values properly, the sum becomes 6666600

Answer 6

Sum of all n digit numbers that can be formed using n non zero digits.