GATE2014 EC-4: GA-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

$Total\ 5\ digit\ \#'s=5\times 4\times 3\times 2\times 1=120$

$First\ \#=13579$

$last\ \#=97531$

$Sum=\dfrac{120}{2}(13579+97531)=6666600$

@`JEET @Lakshman Patel RJIT

Is this the right approach ?

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Yeah it's seems right.

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Yeah this worked here just as a matter of coincidence because, all the no you will get here by using given digits will be in AP, but say if digits were 1,2,5,9 then it won’t work

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This is totally wrong approach. Its the sum for the arithmetic sequence and clearly doesn’t fits the permutation sum.

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@KUSHAGRA गुप्ता,

No, by this approach, we are ignoring the fact where digits gets repeated in the number.Your first number is : 13579, There will come a number say 13599, here 9 gets repeated, which is not allowed as per the question.

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 $6666600$

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.